matsefi/octogon/volumes/octogon_2018_2_proposed... · 812 octogon mathematical magazine, vol. 26,...
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812 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
Proposed problems
PP28069. 27 If λ > 0 then Γ2 (λ) =
�∞R0
tλ−1e−xtdt
��∞R0
tλ−1e−tx dt
�for all
x > 0.
Mihaly Bencze
PP28070. If x, y > 0 then(n−1)!
∞�
0
tn−1e−(x+y)tdt= xn + yn + n!
nPk=1
1
k
�∞�
0
tn−ke−xtdt
��∞�
0
tk−1e−ytdt
� .
Mihaly Bencze
PP28071. If xk > 0 (k = 1, 2, ..., n) and λ > 0 thenP
cyclic
∞R0
tλ−1e−x1x2tdx ≤ Γ2(n)Γ(2n)
nPk=1
∞R0
t2λ−1e−xktdt.
Mihaly Bencze
PP28072. Solve in Z the equation x+y−zy+z−x + y+z−x
z+x−y + z+x−yx+y−z = 3.
Mihaly Bencze
PP28073. Let be G = (a, a+ 1) ∪ (a+ 1,+∞) where a > 0 andlogb ((x ◦ y)− a) = logb (x− a) logb (y − a) when b ∈ (0, 1) ∪ (1,+∞)1). Prove that (G, ◦) is abelian group2). Let be xk > a+ 1 (k = 1, 2, ..., n) and denote {y1, y2, ..., yn} arearrangement of the set {x1, x2, ..., xn} . Prove thatnP
k=1
logb ((xk ◦ yk)− a) ≤nP
k=1
log2b (xk − a) .
Mihaly Bencze
PP28074. Prove thatnP
k=1
1(k−1)!
∞R0
tk−1e−xtdt =�1 + 1
x
�n − 1 for all x > 0.
Mihaly Bencze
27Solution should be mailed to editor until 30.12.2021. No problem is ever permanently
closed. The editor is always pleased to consider for publication new solutions or new in sights
on past problems.
Proposed Problems 813
PP28075. If a, b, c > 0 thenP 1
Γ(a)
∞R0
ta−1e−xtdt+ 3Γ(a+b+c
3 )
∞R0
ta+b+c
3−1e−xtdt ≥ 2
P 1Γ(a+b
2 )
∞R0
ta+b2
−1e−xtdt.
Mihaly Bencze
PP28076. If xij > 0 (i = 1, 2, ..., n) (j = 1, 2, ..., n) and λ > 0 then�nP
i=1
∞R0
tλ−1e−xi1xi2...ximtdt
�2
≤mQj=1
�nP
i=1
∞R0
tλ−1e−xmij tdt
�.
Mihaly Bencze
PP28077. If xk, yk > 0 (k = 1, 2, ..., n) then�nP
k=1
∞R0
tλ−1e−xkyktdt
�2
≤�
nPk=1
∞R0
tλ−1e−x2ktdt
��nP
k=1
∞R0
tλ−1e−y2ktdt
�for all
λ > 0.
Mihaly Bencze
PP28078. Prove thatnP
k=1
�1
Γ(λ)
∞R0
tλ−1e−k(k+1)tdt
� 1λ
= nn+1 for all λ > 0.
Mihaly Bencze
PP28079. In all triangle ABC holdsQ�
1Γ(A)
∞R0
tA−1e−xtdt
�= 1
xπ for all
x > 0.
Mihaly Bencze
PP28080. If xk > 0 (k = 1, 2, ..., n) and λ > 0 thennP
k=1
1Γ(λ)
∞R0
tλ−1e−xktdt ≥ n1−λ
�nP
k=1
xk
�λ
.
Mihaly Bencze
PP28081. Prove thatnP
k=1
1(k−1)!
∞R0
tk−1e−xtdt = xn−1(x−1)xn for all x > 0, x 6= 1.
Mihaly Bencze
814 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP28082. Prove thatnP
k=1
k2(k2+k+1)2
(k6+k3+3k2+3k+1)(k2−k+1)≤ 9n
n+1 .
Mihaly Bencze
PP28083. In all triangle ABC holds
1).P A3+B3
A2+AB+B2 ≥ 2π3 2).
Ptg
�π(A2+AB+B2)9(A2−AB+B2)
�≤ 3
√3
Mihaly Bencze
PP28084. Prove thatnP
k=1
k2+k+1k2−k+1
≤ 3nn+1 .
Mihaly Bencze
PP28085. Prove that maxx,y∈[0,1]
�√sinπx−√
sinπy�2
≥ 2π − 1
2 .
Mihaly Bencze
PP28086. Prove that (2n+ 1)he2+12e (2n)!
i+he2−12e (2n+ 1)!
i= [e (2n+ 1)!]
for all n ∈ N∗, where [·] denote the integer part.
Mihaly Bencze
PP28087. Prove thatnP
k=0
(2k)!
4k(k!)2(2k+1)=
�
4n−1(n!)2(2n+1)π(2n)!
�
4n(n!)2[2n+1][2n]!
where [·] denote
the integer part.
Mihaly Bencze
PP28088. Prove thatnQ
k=1
[ek!]k ≤�2[e(n+1)!]−n−2
n(n+1)
�n(n+1)2
where [·] denote the
integer part.
Mihaly Bencze
PP28089. Prove that
1). 1 + 13! +
15! + ...+ 1
(2n+1)! =
�
e2−12e
(2n+1)!�
(2n+1)!
2). 1 + 12! +
14! + ...+ 1
(2n)! =
�
e2+12e
(2n)!�
(2n)! where [·] denote the integer part.
Mihaly Bencze
Proposed Problems 815
PP28090. Prove that 1!2!3!...n! ≥�(n+1)![en!]
�n+1where [·] denote the integer
part.
Mihaly Bencze
PP28091. Prove thatnP
k=0
k! ≥ (n+1)(n+1)![en!] where [·] denote the integer part.
Mihaly Bencze
PP28092. Prove thatnP
k=0
�1k!
�λ ≥ n1−λ�[e·n!]n!
�λfor all
λ ∈ (−∞, 0] ∪ [1,+∞) where [·] denote the integer part.
Mihaly Bencze
PP28093. Prove thatnP
k=1
(−1)k kn�1 + 1
k
�k �nk
�= (−1)n n!
nPk=1
1k! .
Mihaly Bencze
PP28094. Prove thatnP
i=0
nPj=0
�4i2i
��4n−4i2n−2i
��4j+22j+1
��4n−2j−22n−2j−1
�= 44n−1 − 42n−1
�2nn
�2.
Mihaly Bencze
PP28095. Determine all λ > 0 for whichnP
k=1
k [λ · k!] = [λ (n+ 1)!]− n− 2
where [·] denote the integer part.
Mihaly Bencze
PP28096. Determine all x, y ∈ N such thatnP
k=0
�nk
�x�yn+kxn
�=
�xnn
�y.
Mihaly Bencze
PP28097. Prove thatnP
k=0
�2nk
��nk
��3n+k3n
�=
�3nn
�2.
Mihaly Bencze
816 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP28098. Prove that∞Pk=1
1k2(k+1)(k+2)(k+3)
= 136
�π2 − 49
6
�.
Mihaly Bencze
PP28099. Prove thatnP
k=0
(2n− 2k + 1)�2n+1k
�≥q(n+ 1) (2n+ 1)
�4n2n
�.
Mihaly Bencze
PP28100. Prove thatnQ
k=1
�nk
�2 ≤�
1n
�2nn+1
��n.
Mihaly Bencze
PP28101. Prove thatnP
k=0
�4n+22k+1
�≤r(n+ 1)
�14
�8n+44n+2
�+ 1
4
�4n+22n+1
�+ 1
2
�4n+22n+1
�2�.
Mihaly Bencze
PP28102. Determine all ak ∈ Q+
�k = 0, 1, ...,
�n2
��such that
2n[n2 ]Pk=0
�n2k
��2kk
�ak =
�2nn
�where [·] denote the integer part.
Mihaly Bencze
PP28103. Prove thatP
0≤i<j≤n
�ni
��nj
�= 1
2
�4n −
�2nn
��.
Mihaly Bencze
PP28104. Determine all ak ∈ N (k = 0, 1, 2, ..., 2n) for which2nPk=0
(−1)k�2nk
��4n−k2n
�ak = (−1)n
�2nn
�.
Mihaly Bencze and Marius Dragan
PP28105. Determine all x, y ∈ N for whichnP
k=1
kx (n− k + 1)y =�n+x+yx+y+1
�.
Mihaly Bencze
Proposed Problems 817
PP28106. Determine all ak > 0 (k = 0, 1, ..., n) for whichnP
k=0
ak�nk
�= 3n.
Mihaly Bencze
PP28107. Denote Fk and Lk the kth Fibonacci respective Lucas numbers.
Prove that 2nP
k=0
3kpLkFk+1 ≤ 3n+1Fn+1.
Mihaly Bencze
PP28108. Determine all x ∈ N for whichnP
k=0
�nk
�x=
�nxn
�.
Mihaly Bencze
PP28109. If ak > 0 (k = 1, 2, ..., n) , thenP�
a21a21−a1a2+a22
�λ≤ n for all
λ ∈ [0, 1] .
Mihaly Bencze
PP28110. If a > 1 then
maxx,y∈[0,1]
�√1− 2a cosπx+ a2 −
p1− 2a cosπy + a2
�2≥ 1.
Mihaly Bencze
PP28111. If a, b, c > 0 thenP 4
q(a+b)2
a2+b2+c2+ab+bc−ca≤ 3.
Mihaly Bencze
PP28112. If xk > 0 (k = 1, 2, ..., n) then determine all λ ∈ R for whichP x41+x2
1x22+x4
2
x21+λx1x2+x2
2≥Px1x2.
Mihaly Bencze
PP28113. In all triangle ABC holds�s2 + r2 + 2Rr
� �s2 + r2 + 4Rr
�≥ 4Rr
�5s2 + r2 + 4Rr
�.
Mihaly Bencze
818 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP28114. If x, y ∈ R then
����3
qcosx+ i sinx+
p(cos 2x− cos 3y) + i (sin 2x− sin 3y) +
+ cos y+i sin y3�
cosx+i sinx+√
(cos 2x−cos 3y)+i(sin 2x−sin 3y)
����� ≤ 5 + 2√3
9
Mihaly Bencze
PP28115. In all triangle ABC holds 4R+rs ≥
q2�2− r
R
�.
Mihaly Bencze
PP28116. In all acute triangle ABC holdsP cos(B+A
2 ) sinC cos(B−A2 )
cos(C−A2 ) sinB cos(C+A
2 )= s2+r2−4R2
2(s2−(4R+r)2).
Mihaly Bencze
PP28117. If x ∈ (0, 1) then π3 cosπxsin3 πx
≥ 1x3 + 2
∞Pk=1
1
(x2−k2)32.
Mihaly Bencze
PP28118. If a, b, c > 0 then1R0
eax2+bx+cdx ≤ e
a3+ b
2+c + max
x,y∈[0,1]
�√eax2+bx+c −
√eay2+by+c
�2.
Mihaly Bencze
PP28119. If a, b, c > 0, and b2 > 4ac then
exp
�−
1R0
ln�ax2 + bx+ c
�dx
�+ max
x,y∈[0,1]
�1√
ax2+bx+c− 1√
ay2+by+c
�2
≥
≥ 1√b2−4ac
ln(b+
√b2−4ac)(2a+b−
√b2−4ac)
(b−√b2−4ac)(2a+b+
√b2−4ac)
.
Mihaly Bencze
PP28120. If x, y ∈ [0, 1] then
max�√
x2 + 3x+ 2−py2 + 3y + 2
�2≥ 23
6 − 8e2.
Mihaly Bencze
Proposed Problems 819
PP28121. Prove that maxx,y∈[0,1]
�1√x+1
− 1√y+1
�2≥ ln 2− e
4 .
Mihaly Bencze
PP28122. If 0 < x ≤ y ≤ z < 1 then√1− z lnxy ≤
�√1− x+
√1− y
�ln z.
Mihaly Bencze
PP28123. If x, y ∈ [0, 1] then max�√
ex −√ey�2 ≥ e−√
e.
Mihaly Bencze
PP28124. If x ∈�0, π4
�then (sinx)sinx ≤ (cosx)cosx .
Mihaly Bencze
PP28125. Let be ak > 0 (k = 1, 2, ..., n) and A = 1n
nPk=1
ak, G = n
snQ
k=1
ak,
H = nn�
k=1ak
. Prove that�GA
�√A−H ≥�HA
�√A−G.
Mihaly Bencze
PP28126. If xk ∈�0, π4
�(k = 1, 2, ..., n) , then
nPk=1
ctg2xk ≥ (logb a)2 when
a =nQ
k=1
sinxk and b =nQ
k=1
cosxk.
Mihaly Bencze
PP28127. In all triangle ABC holds
1). s4 − r4 ≥P�r2a − r2
�rbrc
2).(4s4−R2r2)r2
R2 ≥P�h2a − r2
�hbhc
Mihaly Bencze
PP28128. If 2 < a ≤ b thenbRa
(3x2−5x+4) lnxdx
x(x−2)2≤ 2
�b− a+ ln b−2
a−2
�+ 3
2 lna(b−2)b(a−2) − 3 ln bb−2
aa−2 + 12 ln ab ln
ba .
Mihaly Bencze
820 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP28129. If xi > 0 (i = 1, 2, ..., n) andnP
i=1x2i = 1 then
Pcyclic
x2k−11
x2+x3+...+xn≥ 1
nk−2(n−1)for all k ∈ N, k ≥ 1.
Mihaly Bencze
PP28130. In all triangle ABC holds 5P r2a
r2−rra+r2a+ 2
P h2a
r2−rha+h2a≤ 27.
Mihaly Bencze
PP28131. In all triangle ABC holdsP b+c−a
5a2+4b2+4c2−6bc+ab+ca≤ 1
a+b+c .
Mihaly Bencze
PP28132. In all triangle ABC holds
1).P 1
r3a(r2br2c+r2rbrc+4r2(r2b+r2c))≥ 1
2s4r3
2).P 1
h3a(h2
bh2c+r2hbhc+4r2(h2
b+h2
c))≥ R2
8s4r5
Mihaly Bencze
PP28133. Prove thatnP
k=1
1�
sin�
k(k+1)n(n+1)
�2 ≤ n2 + n�1− 4
π2
�.
Mihaly Bencze
PP28134. If x ∈�0, π
2n
�then
nPk=1
1k2
≥ 2n2x2 sinxn sinx−sinnx cos(n+1)x − n
�1− 4
π2
�.
Mihaly Bencze
PP28135. If x ∈�0, π
2n
�then
nPk=1
q1− 4
π2 + 1k2x2 ≥ n2 sin x
2
sin nx2
sin(n+1)x
2
.
Mihaly Bencze
PP28136. If�0, π2
�then 1
sin2 x cos2 x≤ 2
�1− 4
π2
�+ 8x2−4πx+π2
x2(π−2x)2.
Mihaly Bencze
Proposed Problems 821
PP28137. If 0 < a ≤ b ≤ 1 thenbRa
x+1�√
5−3x+2√
(x+1)(4x+5)�2dx ≤ 1
6√5arctg 6(b−a)
√5
45+4ab .
Mihaly Bencze
PP28138. Solve in Z the equation x3+y2z
(y+z)3+ y3+z2x
(z+x)3+ z3+x2y
(x+y)3= 3
4 .
Mihaly Bencze
PP28139. In all triangle ABC holds
1).P�r
2+3 cos2 A2
1+sin2 A2
+ 2q5 + 4 sin2 A
2
�2
≥ 139 + r2−s2
2R2
2).P�r
2+3 sin2 A2
1+cos2 A2
+ 2q5 + 4 cos2 A
2
�2
≥ 143 + 8Rr+r2−s2
2R2
Mihaly Bencze
PP28140. In all triangle ABC holdsP�q
5−3 sinA1+sinA + 2
√5 + 4 sinA
�2
≥ 135 +2(s2−r(4R+r))
R2 .
Mihaly Bencze
PP28141. In all acute triangle ABC holdsP�q
5−3 cosA1+cosA + 2
√5 + 4 cosA
�2
≥ 147− 2(s2−4Rr−r2)R2 .
Mihaly Bencze
PP28142. If ak > 0 (k = 1, 2, ..., n) , thenP
a21�a42 − a22 + 1
�≥ 1
n3
�nP
k=1
ak
�4
.
Mihaly Bencze
PP28143. If x ∈�0, π2
�then�q
5−3 sinx1+sinx + 2
√4 sinx+ 5
�2
+�q
5−3 cosx1+cosx + 2
√4 cosx+ 5
�2
≥ 94.
Mihaly Bencze
822 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP28144. Let (Fn)n≥0 and (Ln)n≥0 be the Fibonacci and the Lucassequence, respectively. Compute the following limits:
a). limn→∞
�2n+2p(n+ 1)!!Fn+1 − 2n
√n!Fn
�√n
b). limn→∞
�2n+2p(n+ 1)!!Ln+1 − 2n
√n!Ln
�√n
D.M. Batinetu-Giurgiu and Neculai Stanciu
PP28145. Let a, b, c ≥ 0. Prove that3√4 + 17a2b+ 3
√4 + 17b2c+ 3
√4 + 17c2a+ 10
�127 − abc
�≥ 5
Paolo Perfetti
PP28146. Let (an)n be a sequence, given by therecurence:man+1 − (m− 2) an + an−1 = 0, where m ∈ R is a parameter andthe first two terms of (an)n are known real numbers. Find m ∈ R, so thatlimn→∞
an = 0.
Laurentiu Modan
PP28147. If a, b, c are real numbers for which ab+ bc+ ca = 1 prove that4a2 + 5b2 + 6c2 > 2
√6
Dorin Marghidanu
PP28148. Let u, v, w complex numbers such that:u+ v + w = 1, u2 + v2 + w2 = 3, uvw = 1Prove that:a). u, v, w are distinct numbers two by two;b). if S(k) := uk + vk + wk, then S(k) is an odd natural number;
c). the expression u2n+1−v2n+1
u−v + v2n+1−w2n+1
v−w + w2n+1−u2n+1
w−u is an integernumber.
Dorin Marghidanu
PP28149. If a, b, c are strictly positive real numbers, prove that:a
a2bc+b4+c4+ b
b2ca+c4+a4+ c
c2ab+a4+b4≤ 1
abc
Dorin Marghidanu
Proposed Problems 823
PP28150. If a, b, c > 0 and m,n, p ∈ (2,+∞) then prove the inequality
mp
ab +
n
qbc +
pp
ca ≥ m+n+p
m+n+p√mmnnpp
Dorin Marghidanu
PP28151. If 1 < a < b < c, prove that√a2 − 1 +
√b2 − 1 +
√c2 − 1 ≤ ab+bc+ca
2 and specify when equality holds.Generalization.
Dorin Marghidanu
PP28152. Prove thata2018
b2018+ b2018
c2018+ c2018
d2018+ d2018
a2018≥ a2017
b2017+ b2017
c2017+ c2017
d2017+ d2017
a2017.
Ovidiu Bagdasar
PP28153. Find the cardinal of the setB = {ij : (i, j) ∈ N× N, 0 ≤ i, j ≤ 2018},
Ovidiu Bagdasar
PP28154. Find the cardinal of the set containing all integers1 ≤ x ≤ 20182018, such that their highest prime factor is 2017.
Ovidiu Bagdasar
PP28155. Let a, b, c be non-negative real numbers. Find all positive real
k such that the inequality kabc+ a2b+ b2c+ c2a ≤ (k + 3) (a+ b+ c)3
27holds
for any nonnegative real a, b.c.
Arkady Alt
PP28156. Prove that inequality
2�xa+
y
b+
z
c
�· x+ y + z
a+ b+ c−�xyab
+yz
bc+
zx
ca
�≥ 3
�x+ y + z
a+ b+ c
�2
holds for
any triangle with sidelengths a, b, c and any positive real x, y, z anddetermine when equality occurs.
Arkady Alt
824 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP28157. Prove that:R π
20
Qnk=1
�sin2k+1 x+ cos2k+1 x
�dx ≥ π
2(√2)n
2 ;n ∈ N∗
Daniel Sitaru
PP28158. If 2 ≤ a ≤ b then 2 ln ba ≥
bRa
3x2+x+4x4+2
dx.
Mihaly Bencze
PP28159. If 2 ≤ a ≤ b then 2 ln b(a+1)a(b+1) ≥
bRa
3x2+x+4(x+1)(x4+2)
dx
Mihaly Bencze
PP28160. In all triangle ABC holds
1).�s2+2Rr+5r2
2
�s≥Q (ma)
a
2).
�s(s2+2Rr+5r2)3(s2−r2−4Rr)
� 32(s
2−r2−4Rr)≥Q
am2a
Mihaly Bencze
PP28161. In all triangle ABC holdsQ
whaa ≥
�r(s2+r2+4Rr)
R+2r
� s2+r2+4Rr4R
.
Mihaly Bencze
PP28162. If x ≥ 2 and λ ≥ 0 then 2x4
λ+5 + 4λ+1 ≥ 3x3
λ+4 + x2
λ+3 + 4xλ+2 .
Mihaly Bencze
PP28163. In all triangle ABC holdsP
a2m4a ≥ s2(s2+2Rr+5r2)
12 .
Mihaly Bencze
PP28164. Solve in R the following system
2�x41 + 2
�= 3x32 + x23 + 4x4
2�x42 + 2
�= 3x33 + x24 + 4x5
−−−−−−−−−−−−2�x4n + 2
�= 3x31 + x22 + 4x3
Mihaly Bencze
Proposed Problems 825
PP28165. If x ≥ 2 then48
�x4 + 2
� �x4 + 10
�≥ 5x2
�3x2 + x+ 4
� �9x2 + 4x+ 24
�.
Mihaly Bencze
PP28166. In all triangle ABC holds
1). 7Pq
73 + 2 sin2 A2 ≥ 201− s2+8Rr−r2
R2
2). 7Pq
73 + 2 cos2 A2 ≥ 183 + s2−r2
R2
Mihaly Bencze
PP28167. If x ∈�0, π2
�then
7�√
73 + 2 sinx+√73 + 2 cosx
�≥ 110 + 16 (sinx+ cosx) .
Mihaly Bencze
PP28168. In all triangle ABC holds:
1). 7P√
73 + 2 sinA ≥ 177 +4(4sR−s2+r(4R+r))
R2
2). 7P√
73 + 2 cosA ≥ 177 +4(s2−6R2−3Rr−r2)
R2
Mihaly Bencze
PP28169. If ak > 0 (k = 1, 2, .., n) thennP
k=1
(2a3k+1)2
ak≥ 1
n
�n+
nPk=1
ak +nP
k=1
a2k
�2
.
Mihaly Bencze
PP28170. Prove thateR1
q1 + 1
x2dx >√e2 − 2e+ 2.
Mihaly Bencze
PP28171. In all triangle ABC holds 3 (2R+ r)2 + 2r2 ≥ s2 + 8R2.
Mihaly Bencze
PP28172. In all triangle ABC holdsP
(ma)λ ≥ 3
�2r(s2−r(4R+r))
3R2
�λ
for all
λ ∈ (−∞, 0] ∪ [1,+∞) .
Mihaly Bencze
826 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP28173. If ak > 0 (k = 1, 2, ..., n) andnP
k=1
ak = 1 then
nPk=1
�ak − 1
2
�3 ≥ 3(2−n)3
8n3 .
Mihaly Bencze
PP28174. In all triangle ABC holdsP �
a cos B−C2
�λ ≥ 3�s(R+2r)
3R
�λfor all
λ ∈ (−∞, 0] ∪ [1,+∞) .
Mihaly Bencze
PP28175. If x ∈�0, π2
�then
4π
�arctg
�12 sin
2 x�+ arctg
�12 cos
2 x��
+ 59100 > 8
14−sin2 x cos2 x.
Mihaly Bencze
PP28176. In all triangle ABC holdsP a2 cos2(B−C)
sin 2B+sin 2C = 4sr.
Mihaly Bencze
PP28177. In all triangle ABC holds�
a3 cosA�
a cos3 A=
4R2(s2−r2−4Rr−3R2)5R2−s2+r2+4Rr
.
Mihaly Bencze
PP28178. In all acute triangle ABC holds
9 ≤ (R+r)(s2+r2−4R2)R(s2−(2R+r)2)
≤ 9 +P �� 1
cosA − 1cosB
�� .
Mihaly Bencze
PP28179. In all triangle ABC holds
1). 9 ≤ (2R−r)(s2+r2−8Rr)2Rr2
≤ 9 +P ����
1sin2 A
2
− 1sin2 B
2
����
2). 9 ≤ (4R+r)(s2+(4R+r)2)2Rs2
≤ 9 +P ����
1cos2 A
2
− 1cos2 B
2
����
Mihaly Bencze
PP28180. Let a, b ∈ N , determine all n ∈ N for which a2n+ b2 andb2n+ a2 are both perfect squares.
Mihaly Bencze
Proposed Problems 827
PP28181. In all triangle ABC holds 9 ≤ s2+r2+4Rr2Rr ≤ 9 +
P �� 1sinA − 1
sinB
�� .
Mihaly Bencze
PP28182. In all triangle ABC holdsP
AH3 ≥ 12(4+3√3)(sr−R2)R11 .
Mihaly Bencze
PP28183. In all triangle ABC holds 1r
P 1rarb
≤ 1rarbrc
+ 827r3
.
Mihaly Bencze
PP28184. In all triangle ABC holdsP a(a2+bc)
(b+c)2≤ 4s(s2−r2−Rr)
9(s2+r2+2Rr)
P a2+bca(b+c) .
Mihaly Bencze
PP28185. In all triangle ABC holdsP�
ara
�λ≥ 3
�2(4R+r)
3s
�λfor all
λ ∈ (−∞, 0] ∪ [1,+∞) .
Mihaly Bencze
PP28186. In all triangle ABC holds�P a
ra
�2+�
�
a�
ra
�2≥ 8.
Mihaly Bencze
PP28187. Determine all xk ≥ 3 (k = 1, 2, ..., n) for which
nQk=1
(3xk − 1)2xk ≤
3n�
k=1xk−n
n�
k=1xk−n
n
.
Mihaly Bencze
PP28188. Solve in R the equation (3x + 1)y + 2 · 3x = 3z + 1.
Mihaly Bencze
PP28189. If a, b, c, d ∈ (0,∞), prove that:ln a2 + ln b
4 + ln c6 + ln d
12 ≤ ln�a2 + b
4 + c6 + d
12
�.
Dorin Marghidanu
828 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP28190. Solve the equation tg πx tg
π2x =
√x− 2
Ionel Tudor
PP28191. Prove that there are only finitely many n, k, p ∈ N∗ such thatnQ
r=1
�nk
rp + rp�∈ N.
Mihaly Bencze
PP28192. If ak ∈ R (k = 1, 2, ..., n) such thatP a1
a2+a3+...+an= 1 then
determine all r ∈ N for which�P (a2+a3+...+an)
r
a2r1
� 1r ∈ Q.
Mihaly Bencze
PP28193. Find all n, k ∈ N for which an − 1 is divisible by knk for alla ∈ N, (a, n) = 1.
Mihaly Bencze
PP28194. Let M be the set of all rational numbers expressible in the formnQ
k=1
a2k−ak+1
b2k−bk+1
where ak, bk ∈ N (k = 1, 2, ..., n) . Prove that M contain
infinitely many prime numbers.
Mihaly Bencze
PP28195. If x, y, z > 0 and x+ y + z = 12λ then
2λ2P 1
3λ(x+y)+1 ≤ 1�
(x+y)(y+z) ≤λ3
P 1x+y .
Mihaly Bencze
PP28196. If a, b, c > 0 andP
a = 3 thenP
a2 ≥ 3 + 19 (P |a− b|)2 .
Mihaly Bencze
PP28197. In all triangle ABC holdsP hλ
b+hλ
c
rλa≥ 6 for all
λ ∈ (−∞, 0] ∪ [1,+∞) .
Mihaly Bencze
Proposed Problems 829
PP28198. Denote Fk the kth Fibonacci number. Prove that�nP
k=1
n−k+1FkFk+2
��nP
k=1
1Fk+1Fk+2
�≤
�n2
�2.
Mihaly Bencze
PP28199. In all triangle ABC holdsn√r2 +
Pnp
r2a +P n
√a2 ≤ n
q7n−1 (4R)2, for all n ∈ N∗.
Mihaly Bencze
PP28200. In all triangle ABC holdsQ
(ha + hb) ≤ 8s2r.
Mihaly Bencze
PP28201. Let A1A2...An be a convex polyhedron and B1B2...Bn a regularpolyhedron, both inscribed in the same sphere. Prove thatP1≤i<j≤n
(BiBj −AiAj) ≥ 0.
Mihaly Bencze
PP28202. In all triangle ABC holdsP n
√a ·AI2 ≤ n
√3n−1abc for all
n ∈ N∗.
Mihaly Bencze
PP28203. Let A1A2...An be a convex polygon and B1B2...Bn a regularpolygon, both inscribed in the same circle. Prove thatP1≤i<j≤n
(BiBj −AiAj) ≥ 0.
Mihaly Bencze
PP28204. Prove thatnP
k=1
5
r(k14+k7+1)2
3(k18+k9+1)≥ n(n+1)
2 .
Mihaly Bencze
PP28205. Let ABC be a triangle, and M ∈ Int (ABC) . The linesAM,BM,CM cut the circumcircle in points D,E and F . Prove thatAB ·BC · CA ·DE · EF · FD ≥ 27BD2 · CE2 ·AF 2.
Mihaly Bencze
830 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP28206. If x > 0 then�x14 + x7 + 1
�≥ 3x5
�x18 + x9 + 1
�.
Mihaly Bencze
PP28207. In all triangle ABC holds r2λ +P
r2λa +P
a2λ ≥ 71−λ (4R)2λ forall λ ∈ (−∞, 0] ∪ [1,+∞) .
Mihaly Bencze
PP28208. In all triangle ABC holdsP�
ha+hb
rc
�λ≥ 3 · 2λ for all
λ ∈ (−∞, 0] ∪ [1,+∞) .
Mihaly Bencze
PP28209. In all triangle ABC holds (2Rr)s ≥ (AI)a (BI)b (CI)c .
Mihaly Bencze
PP28210. In all triangle ABC holdsP
(IA)λ +P
(IaA)λ ≥ 6 (abc)λ3 for all
λ ≥ 0.
Mihaly Bencze
PP28211. Let ABC be a triangle. Determine all point M∈ Int (ABC) forwhich aMA2 + bMB2 + cMC2 = abc.
Mihaly Bencze
PP28212. In all triangle ABC holdsP
IA4 ≥ (abc)2�
a2.
Mihaly Bencze
PP28213. If x, y, z > 0 then 3xyz +P
x2 (y + z) ≤ (P
x)�P
x2�.
Mihaly Bencze
PP28214. If x, y, z > 0 then (P
x)2 (P
xy)2 ≥ 4xyz (x+ y) (y + z) (z + x) .
Mihaly Bencze
PP28215. If x, y, z > 0 then (P
x) (P
xy) ≤ 98
Q(x+ y) .
Mihaly Bencze
Proposed Problems 831
PP28216. If x, y, z > 0 then (P
x) (P
xy) ≥ 3xyz + 13 (P
x√y + z)2 .
Mihaly Bencze
PP28217. If x, y, z > 0 then (P
x) (P
xy) ≤ 3xyz + 127
P(2x+ y + z)3
Mihaly Bencze
PP28218. Solve in N the equationnP
k=1
xtk + n− 1 = LCM�xt1, x
t2, ..., x
tn
�.
Mihaly Bencze
PP28219. Let A1A2...An be a polygon circumscribed to the circle withradius r and I the incenter, and denote G his centroid,λ = min |AiAj −AkAp| , i 6= j, k 6= p, i 6= k, j 6= p, i, j, k, p ∈ {1, 2, ..., n} .Prove that
nPk=1
Area [AkIG] ≥ (n−1)λrn .
Mihaly Bencze
PP28220. Solve in C the following system:
6x− 2y + 9z2 − 12z = −8x2 − 4y2 + 2x+ 4y − 6z = −46x2 − 12y2 − 18z2 + 12x+ 10y + 24z = 4
Mihaly Bencze
PP28221. Let be f : N → N such that f (3k) = k, f (3k + 1) = 3k + 2,f (3k + 2) = 3k + 3. Prove that for each n ∈ N, there exist k ∈ N such thatf ◦ f ◦ ... ◦ f| {z }
k−time
(n) = 1.
Mihaly Bencze
PP28222. Determine all pentagons A1A2A3A4A5 in whichA2A2
5A1A2
= A1A2 +A2A5.
Mihaly Bencze
PP28223. In all triangle ABC holdsP tgA
1+tg2A = srR2 .
Mihaly Bencze
832 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP28224. Determine all r, s ∈ N for which (nr)!n−1Qk=1
k!(ks+n)! ∈ N.
Mihaly Bencze
PP28225. If a, b, c > 0 thenP a+3b+2c
3a+5b+4c ≤ 32 .
Mihaly Bencze
PP28226. If a, b, c > 0 thenP a2
3a+ 3√
9(a+b+c)bc≥ a+b+c
6 .
Mihaly Bencze
PP28227. If a, b > 0 then�a√b+b
√a√
a+b+ a√
2
���√a+
√b�√
a+ b+ a√2�+
+�a√b+b
√a√
a+b+ b√
2
���√a+
√b�√
a+ b+ b√2�≤ 5a2 + 8ab+ 5b2.
Mihaly Bencze
PP28228. In all triangle ABC holds�4R+rsr
�2 ≥ 2r
�2r − 1
R
�.
Mihaly Bencze
PP28229. Prove that there is a constant d > 0 with the following property:If a, b, c, n ∈ N∗ such that gcd (a+ i, b+ j, c+ k) > 1 for alli, j, k ∈ {1, 2, ..., n} then min {a, b, c} > (dn)n .
Mihaly Bencze
PP28230. If xk > 0 (k = 1, 2, ..., n) , then P
cyclic
1x2+x3+...+xn
! P
1≤i<j≤nxixj
!≤ n
2
nPk=1
xk.
Mihaly Bencze
PP28231. Find all functions f : R → R such that1 + f (x) f (y) f (z) = f (x+ y + z) + f (xy + yz + zx) + xyz for allx, y, z ∈ R.
Mihaly Bencze
Proposed Problems 833
PP28232. If ak > 0 (k = 1, 2, ..., n) , then
Pcyclic
a41+4a31a2+6a21a22+4a1a32+a42
5a21+2a1a2+5a22≥ 2
Pcyclic
a1a2
Mihaly Bencze
PP28233. Solve in Z the equation x5
x+4y + y5
y+4z + z5
z+4x = 35 .
Mihaly Bencze
PP28234. In all triangle ABC holds
1).P 5a2+2ab+5b2
a4+4a3b+6a2b2+4ab3+b4≤ 1
4Rr
2).P 5h2
a+2hahb+5h2b
h4a+4h3
ahb+6h2ah
2b+4hah3
b+h4
b
≤ s2+r2+4Rr8s2r2
3).P 5r2a+2rarb+5r2
b
r4a+4r3arb+6r2ar2b+4rar3b+r4
b
≤ (4R+r)2−2s2
s4r2
Mihaly Bencze
PP28235. In all triangle ABC holdsP
mλa ≥ 3
�2r(s2−4Rr−r2)
3R2
�λ
for all
λ ∈ (−∞, 0] ∪ [1,+∞) .
Mihaly Bencze
PP28236. If a, b > 0 then(2a+b)(a2+2b2)
2a2+b2+
(2b+a)(b2+2a2)2b2+a2
≥ 3 (a+ b) .
Mihaly Bencze
PP28237. Prove thatnP
k=1
k
|2k√k2+1−2k2−1| ≥ n2 (n+ 1)2 .
Mihaly Bencze
PP28238. In all triangle ABC holds
Pq(s−a)(s−b)
ac ≥ 14
�s2
s2−2r2−8Rr
�2− s2+r2−8Rr
8Rr .
Mihaly Bencze
834 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP28239. Solve in R the following system:
x31x2 = (x3 + x4)3
x32x3 = (x4 + x5)3
−−−−−−−−x3nx1 = (x2 + x3)
3
.
Mihaly Bencze
PP28240. Let be a ∈ N∗. Prove thatnP
k=1
��a− k (k + 1)√2�� ≥ (
√3−
√2)n
n+1 .
Mihaly Bencze
PP28241. In all triangle ABC holds:
1).P 1
cos2 A2+sin4 A
2
≤ 108R2
28R2+2Rr+r2
2).P 1
sin2 A2+cos4 A
2
≤ 108R2
28R2+2Rr+r2
Mihaly Bencze
PP28242. Denote Fk, Lk, Pk the kth Fibonacci, Lucas, Pell numbers, and
xn =
Fn if n = 3kLn if n = 3k + 1Pn if n = 3k + 2
. Compute∞Pk=1
1x2k
.
Mihaly Bencze
PP28243. In all triangle ABC holds
1).P 1
1−sin2 A2+sin4 A
2
+ 108R2
28R2+2Rr+r2≤ 8
P 1
2(cos2 A2+cos2 B
2 )+(sin2 A
2+sin2 B
2 )2
2).P 1
1−cos2 A2+cos4 A
2
+ 108R2
28R2+2Rr+r2≤ 8
P 1
2(sin2 A2+sin2 B
2 )+(cos2A2+cos2 B
2 )2
Mihaly Bencze
PP28244. In all acute triangle ABC holdsP 11−cosA+cos2 A
+ 27R2
7R2−Rr+r2≤ 8
P 14−2(cosA+cosB)+(cosA+cosB)2
.
Mihaly Bencze
PP28245. Solve in Z the equation x3
x+2y + y3
y+2z + z3
z+2x = 1.
Mihaly Bencze
Proposed Problems 835
PP28246. Let A1A2A3A4A5 be a pentagon with sides a1, a2, a3, a4, a5. Thediagonals of pentagon A1A2A3A4A5 determine the pentagon B1B2B3B4B5
with sides b1, b2, b3, b4, b5. Prove thatP
cyclic
�a1
a2+a3+ b1
b2+b3
�≥ 5.
Mihaly Bencze
PP28247. In all triangle ABC holdsP 1
1−sinA+sin2 A≤ 27R2
s2−3sR+9R2 .
Mihaly Bencze
PP28248. In all acute triangle ABC holdsP 1
1−cosA+cos2 A≤ 27R2
7R2−Rr+r2.
Mihaly Bencze
PP28249. In all triangle ABC holds1).
P 11−sin4 A
2+sin8 A
2
+ 1728R4
(8R2+r2−s2)2−24R2(r2−s2)+320R4≤
8P 1
4−2(sin4 A2+sin4 B
2 )+(sin4 A
2+sin4 B
2 )2
2).P 1
1−cos4 A2+cos8 A
2
+ 1728R4
((4R+r)2−s2)(8Rr−s2−8R2)+576R4≤
8P 1
4−2(cos4 A2+cos4 B
2 )+(cos4A2+cos4 B
2 )2
Mihaly Bencze
PP28250. If 0 < a ≤ b < π2 then ctga−ctgb
b−a ≤ 1ab + 1− 4
π2 .
Mihaly Bencze
PP28251. If x ∈�0, π2
�then 8
π2 + 1sin2 x cos2 x
≤ 1x2+ 1
(π2 −x)2+ 2.
Mihaly Bencze
PP28252. In all acute triangle ABC holds12π2 +
�s2+r2+Rr
2sr
�2≤ 3 +
P 1A2 + 4R
r .
Mihaly Bencze
PP28253. In all acute triangle ABC holds�s2+r2−4R2
s2−(2R+r)2
�2+ 12
π2 ≤ 3 +P 1
(π2−A)
2 + 8R(R+r)
s2−(2R+r)2.
Mihaly Bencze
836 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP28254. In all triangle ABC holds
1).�sr
�2+ 12
π2 ≤ 2 + 4P 1
A2 + 8Rr
2).�4R+r
s
�2+ 12
π2 ≤ 2 + 4P 1
(π2−A)
2
Mihaly Bencze
PP28255. In all triangle ABC holds
1).Q �
4A2 + 1− 4
π2
�≥
�4Rr
�2
2).Q�
4
(π2−A)
2 + 1− 4π2
�≥
�4Rs
�2
Mihaly Bencze
PP28256. If x, y, z ≥ 1 thenP 1x2−x+1
+ 27(�
x)2−3�
x−19≥ 8
P 1(x+y)2−2(x+y)+4
. If x, y, z ∈ (0, 1) then
holds the reverse inequality.
Mihaly Bencze
PP28257. In all triangle ABC holdsP 11−sinA+sin2 A
+ 27R2
9R2−3sR+s2≤ 8
P 14−2(sinA+sinB)+(sinA+sinB)2
.
Mihaly Bencze
PP28258. Prove that
1).nP
k=1
1k2−k+1
≥ 4nn2+3
2).nP
k=1
1k4−k2+1
≥ 36n4n4+12n3+n2−12n+31
Mihaly Bencze
PP28259. Prove thatnP
k=1
k2(k+1)2
k4+2k3−k+1≤ n(n+1)2
n2+n+1.
Mihaly Bencze
PP28260. If xk ≥ 1 (k = 1, 2, ..., n) thennP
k=1
1x2k−xk+1
≥ n3�
n�
k=1xk
�2
−nn�
k=1xk+n2
.
Mihaly Bencze
Proposed Problems 837
PP28261. Denote Fk and Lk the kth Fibonacci respective Lucas numbers.Prove that:
1).nP
k=1
1F 2k−Fk+1
≥ n3
F 2n+2−(n+2)Fn+2+n2+n+1
2).nP
k=1
1L2k−Lk+1
≥ n3
L2n+2−(n+6)Ln+2+n2+3n+9
Mihaly Bencze
PP28262. If a, b, c > 0 and a+ b+ c ≥ 12 then
P(a+ b)4 + 7
P �ab+ bc+ ca+ b2
�2 ≥p8Q
(a2 + 2b2 + c2 + 2 (a+ c) b).
Mihaly Bencze
PP28263. Prove thatnP
k=1
1
(nk)2−(nk)+1
≥ (n+1)3
4n−(n+1)2n+(n+1)2.
Mihaly Bencze
PP28264. In all acute triangle ABC holds 12P tgA
2A ≥ 18 +
PAtgA
2 .
Mihaly Bencze
PP28265. If x, y, z > 0 and x+ y + z = 1λ+1 where λ > 0 thenP x+λy+1
(x+λy)3+1≤ 27
7 .
Mihaly Bencze
PP28266. If ak ∈ R (k = 1, 2, ..., n) such that |ai − aj | ≥ πi−j for all i > j.Find the minimal value of
P1≤i<j≤n
(ai − aj)2 .
Mihaly Bencze
PP28267. If x, y, z > 0 then
Pp(x+ y) (y + z) (x+ 2y + z) ≥
pQ(x+ 2y + z)+2 (
Px)q
2�
(x+y)3�
x2+9�
xy.
Mihaly Bencze
838 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP28268. If xk > 0 (k = 1, 2, ..., n) and
�nP
k=1
x2k
��nP
k=1
xk
�2
= 1 then
Pcyclic
x51
x2+x3+...+xn≥ 1
n(n−1)
�
n�
k=1xk
�4 .
Mihaly Bencze
PP28269. If Fk denote the nth Fibonacci number, thennP
k=1
�Fk
Fk+Fk+2+
Fk+1
Fk+1+Fk+3− Fk+Fk+2
Fk+2Fk+2
�≤ 0.
Mihaly Bencze
PP28270. If a, b, c > 0 and a+ b+ c = 12 then
P a+b2+8(a2+b2)+10ab+13c
≥ 14 .
Mihaly Bencze
PP28271. Let be ak ∈ N∗ (k = 1, 2, ..., n) . Prove thatP
cyclic
lcm(a1,a2)+gcd(a1,a2)a1+a2
≤ 12
nPk=1
ak.
Mihaly Bencze
PP28272. Prove that if there exist sets M1,M2, ...,Mm satisfying thefollowing conditions:
1). For all 1 ≤ i ≤ m, Mi ⊆ {F1, F2, ..., Fn}2). For all 1 ≤ i < j ≤ m, min {|Fi − Fj | , |Fj − Fi|} = 1then m ≤ n, where Fk denote the kth Fibonacci number.
Mihaly Bencze
PP28273. In all triangle ABC holdsP �
ctgA2 ctg
B2 − tgA
2 tgB2
�2 ≥ 4.
Mihaly Bencze
PP28274. Let ABC be a triangle, and D ∈ Int (ABC) such that I, B,C,Dare concyclic. The line through C parallel to BD meets AD at point E.Determine all triangles ABC in which CD2 = BD · CE.
Mihaly Bencze
Proposed Problems 839
PP28275. If a > 0 then solve in R the equationax + 2a2x + ...+ nanx + n(n−1)
2 = n2ax.
Mihaly Bencze
PP28276. Let A,B ∈ Mn (C) . Determine all x, y ∈ R for which(A+ xB)A∗ (A− yB) = (A− xB)A∗ (A+ yB) .
Mihaly Bencze
PP28277. In all triangle ABC holdsP sin 7A
sinA − 8Q
cos 2A = 5 + 2rR − 2(s2−(2R+r)2)
R2 .
Mihaly Bencze
PP28278. In all triangle ABC holds
knm�
a2
k+1 + b2
n+1 + c2
m+1
�≥
≥ 4√3Area [ABC]
kPu=1
nPv=1
mPw=1
q1
u(u+1)v(v+1) +1
v(v+1)w(w+1) +1
u(u+1)w(w+1) .
Mihaly Bencze
PP28279. Let be A,B ∈ Mn (C). Determine all x, y ∈ Z such thatdet (In +AxByA∗) = det (In +A∗BxAy) .
Mihaly Bencze
PP28280. Prove that the equation x3 + 3x2 + (3− 2n)x− n+ 1 = 0 havethree real roots x1 (n) < x2 (n) < x3 (n) and compute lim
n→∞x2 (n) .
Mihaly Bencze
PP28281. In all triangle ABC holdsPcos (sinA) +
Psin (cos (tgA)) > 3
p2−
√2.
Mihaly Bencze
PP28282. Determine all injective functions f : R → R for whichnP
k=1
f (axk) =nQ
k=1
ak when ak > 0 (k = 1, 2, ..., n) .
Mihaly Bencze
840 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP28283. Let ABC be a triangle in which b2 + c2 = 5a2. Prove that
a�
12c+
√10a
+ 12b+
√10a
�+ b
5a+√10a
+ c5a+
√10b
= 2ab+c .
Mihaly Bencze
PP28284. Determine all n ∈ N for which�112n − 26n
� �n4 − 1
�is divisible
by 285.
Mihaly Bencze
PP28285. If x ∈�0, π2
�then
tg (cos (sinx)) + tg (sin (cos (tgx))) > 2tg�π2
p2−
√2�.
Mihaly Bencze
PP28286. Let ABC be a triangle in which A∡ = 90◦. Prove that
1wa(wa+BD)CD + 1
wa(wa+CD)BD + 1(wa+BD)CD2 + 1
(wa+CD)BD2 = 4aw2
a.
Mihaly Bencze
PP28287. If x, y, z ∈ [0, 1] then
x+ y + z ≤r�
1−√1− x2
��1 +
p1− y2
��1 +
√1− z2
�+
+
r�1 +
√1− x2
��1−
p1− y2
��1 +
√1− z2
�+
+
r�1−
√1− x2
��1−
p1− y2
��1 +
√1− z2
�.
Mihaly Bencze
PP28288. Prove that the equation {xn}+ {yn} = {zn} (n ∈ N,n ≥ 2) havean infinitely many solutions in Q\Z, when {·} denote the fractional part.
Mihaly Bencze
PP28289. Determine all differentiable functions f : R → R for which
f ′ �ax3 + bx�= c and f (0) = 0 and
2R0
f (x) dx = 197210 .
Mihaly Bencze
Proposed Problems 841
PP28290. Determine all n ∈ N∗ for whichcard
�nx ∈ N |x3−x2−x−1
n ∈ No∩ {1, 2, 3, ..., n}
�= 1.
Mihaly Bencze
PP28291. Let be f : R → [0,+∞) a continuous function with period T > 0and gk : [0, T ] → R continuous function (k = 1, 2, ..., n) . Prove that
limn→∞
TR0
f (nx)nQ
k=1
gk (x) dx = 1TN−1
TR0
f (x) dx
! nQ
k=1
TR0
gk (x) dx
!.
Mihaly Bencze
PP28292. Let be 0 < x1 < x2 < x3, α,β, γ ∈ (0, 1) and x3n+1 = xα3n−1x1−α3n ;
x3n+2 = xβ3nx1−β3n+1, x3n+3 = xγ3n+1x
1−γ3n+2 for all n ≥ 1. Prove that the sequence
(xn)n≥1 is convergent.
Mihaly Bencze
PP28293. Let ABCD be a convex quadrilateral in which A∡ = C∡ = 90◦.Prove that 4
BD2 = 1(AD+AB)2
+ 1(AD−AB)2
+ 1(CD+CB)2
+ 1(CD−CB)2
.
Mihaly Bencze
PP28294. Let be A ∈ Mn (R) , A = (aij)1≤i,j≤n . Prove that 1n
nPi,j=1
a2ij
!n
≥ det2 (A) .
Mihaly Bencze
PP28295. Let be G = (a, b) and x ∗ y = (a+b)xy−2ab(x+y)+2ab(a+b)2xy−(a+b)(x+y)+c . Determine
all c ∈ R for which (G, ∗) is abelian group. Prove that (G, ∗) ∼=�R∗
+, ·�.
Mihaly Bencze
PP28296. If a, b, c ∈ C∗ and |a| = |b| = |c| then P���a
b − ba
���2 ≤ 6√3.
Mihaly Bencze
PP28297. In all triangle ABC holds1).
P ab(a+c)2(b+c)2
≤ 132Rr
2).P hahb
(ha+hc)2(hb+hc)
2 ≤ s2+r2+4Rr32R 3).
P rarb(ra+rc)
2(rb+rc)2 ≤ 4R+r
16s2r
4).P sin2 A
2sin2 B
2
(sin2 A2+sin2 C
2 )(sin2 B
2+sin2 C
2 )≤ R(2R−r)
2r2
842 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
5).P cos2 A
2cos2 B
2
(cos2 A2+cos2 C
2 )(cos2B2+cos2 C
2 )≤ R(4R+r)
2s2
Mihaly Bencze
PP28298. Let be p = 3k + 2, k ∈ N∗ and denote an the sum of decimals of
the number pn. ComputenP
k=1
1a2k
.
Mihaly Bencze
PP28299. Prove that if���xn+1
xn+ xm+1
xm− 2xm+n
xmxn
��� < 2m+n for all n,m ∈ N∗
then (xn)n≥1 is a geometrical progression.
Mihaly Bencze
PP28300. In all triangle ABC holdsP
a�AI2 +AH2
�≥ 8sRr.
Mihaly Bencze
PP28301. In all triangle ABC holdsP �
AI4 +AH4�≥ 16s2R2r2
s2−r2−4Rr.
Mihaly Bencze
PP28302. In all acute triangle ABC holdsP
sin (cosA) ≤P sin A2 .
Mihaly Bencze
PP28303. Solve in R the following system
(arcsin x
x+1 + arcsin 3y3y+1 + arcsin 14z
15z+5 = π
2arctg 1x+5 + arctg 1
y = arctg 2z2
2z2+2z+3
Mihaly Bencze
PP28304. Let ak ∈ Z (k = 1, 2, ..., 2n+ 1) distinct numbers such that2n+1Pk=1
|ak| = n (n+ 1) . Prove that
����2n+1Pk=1
a2n+1k
���� = 0.
Mihaly Bencze
PP28305. Prove thatnP
k=1
4
q1k4
+ 14k5
+ 16k6
+ 14k7
+ 1k8
≥ 4√128nn+1 .
Mihaly Bencze
Proposed Problems 843
PP28306. Solve in R the following system:
x2
|y+1| =y−4z+1 + ||z|− 1|
y2
|z+1| =z−4x+1 + ||x|− 1|
z2
|x+1| =x−4y+1 + ||y|− 1|
.
Mihaly Bencze
PP28307. In all triangle ABC holdsP�
w3a
ha
�λ≤
�sR2r
√s2 − 12Rr
�λfor all
λ ∈ [0, 1] .
Mihaly Bencze
PP28308. Solve in R the equationnP
i=1
�n+1n − xi
�k+
�nP
i=1xi
�k
= n+ 1
when k ≥ 2, k ∈ N is given.
Mihaly Bencze
PP28309. If x ≥ 1 then x2n+1
2n+2 + (2n+1)xn
n+1 ≥ (2n+1)xn+1
n+2 + 1 when n ∈ N.
Mihaly Bencze
PP28310. In all triangle ABC holds Rr − 2 ≥ 3
4
��|√a−
√b|
�√a
�2
. A new
refinement of Euler’s R ≥ 2r inequality.
Mihaly Bencze
PP28311. Let be P (x) =nP
k=0
akxk and ε = cos 2π
n+2 + i sin 2πn+2 . Determine
all ak ∈ R (k = 0, 1, ..., n) such thatn+1Qk=1
P�εk�= (n+ 2)n .
Mihaly Bencze
PP28312. Solve in Q the equationnP
k=1
akk! =
1n! .
Mihaly Bencze
844 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP28313. If a ∈ (0, 1) ∪ (1,+∞) thennP
k=1
�ak + a−k
�≥ n(n+1)(2n+1)(ln a)2
6 + 2n.
Mihaly Bencze
PP28314. Let be M =�A ∈ M3 (C) |A2 = I3
. Determine all subgroups of
(M3 (C) , ·) from M.
Mihaly Bencze
PP28315. Let ABC be an acute triangle and xn+1 = xn sin2A+ sin2 B
xn,
yn+1 = yn sin2B + sin2 C
yn, zn+1 = zn sin
2C + sin2 Azn
for all n ∈ N,
x0, y0, z0 > 0. Prove that limn→∞
(xn + yn + zn) ≤s(s2+r2−4R2)
3R(s2−(2R+r)2).
Mihaly Bencze
PP28316. Let ABC be a triangle. Determine maxPq
aa sin2 B+b cos2 C
.
Mihaly Bencze
PP28317. Solve in R the following system:
|x+ y|+ 2z = 22x+ |y − 2z| = 16|x+ 3z|+ y = 13
.
Mihaly Bencze
PP28318. Solve in C the following system:
|x+ 2y|+ 3 |z| =√41 + 3
√34
|x|+ |3y − z| =√5 +
√205
|2x+ z|− 2 |y| =√41− 2
√13
.
Mihaly Bencze
PP28319. In all triangle ABC holds
s (P
AI)3 ≤ 4�s2 − 3r2 − 6Rr
�2P �cos A
2
�3.
Mihaly Bencze
PP28320. Denote Fk respective Lk the kth Fibonacci respective Lucas
numbers. Prove that n�n2 − 1
� nPk=1
FkLk ≤�n2 − 1
�(Fn+2 − 1) (Ln+2 − 3)+
Proposed Problems 845
+3
�nP
k=1
(2k − n− 1)Fk
��nP
k=1
(2k − n− 1)Lk
�.
Mihaly Bencze
PP28321. Determine the gometrical locus of points M in the plane of theconvex quadrilateral ABCD, such that MB = 2MA, MC = 3MA,MD = 4MA.
Mihaly Bencze
PP28322. Determine all a, b, c ∈ N for which�13a + 7b − 2
� �13b + 7c − 2
�(13c + 7a − 2) is divisible by 729.
Mihaly Bencze
PP28323. Determine all a1, a2, a3 > 0 for which
�nP
k=1
ak
�= 1 for all n ≥ 3
where (n+ 1) an+1 = (n− 1) an for all n ≥ 2, and [·] denote the integer part.
Mihaly Bencze
PP28324. Prove that the triangle ABC is equilateral if and only if�ha
wa
�2+
�hb
wb
�2+ 4r
R = 2q
2rR
�ha
wa+ hb
wb
�.
Mihaly Bencze
PP28325. In all triangle ABC holds 3(4R+r)2s2
≥ (4R+r)2+s2
4s2R≥ 6
5R+2r .
Mihaly Bencze
PP28326. Prove thatnP
k=1
arctgkk2+2k+2
< 12
�arctg2 (n+ 1)− π2
16
�<
nPk=1
arctg(k+1)k2+1
.
Mihaly Bencze
PP28327. In all triangle ABC holds16(s2−r2−Rr)9(s2+r2+2Rr)
+ Rrs2−3r2−6Rr
≥ 32 .
Mihaly Bencze
PP28328. In all triangle ABC holds s2+r2−8Rr9Rr + r2
4(s2−12Rr)≥ 3
4 .
Mihaly Bencze
846 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP28329. If a, b, c > 0 and λ > 0 thenP a+(λ+1)b+2λc
(λ+2)a+(2λ+3)b+(3λ+1)c ≤ 32 .
Mihaly Bencze
PP28330. If a, b > 0 then(2a+b)(a2+2b2)
2a2+b2+
(2b+a)(b2+2a2)2b2+a2
≥ 3 (a+ b) .
Mihaly Bencze
PP28331. In all triangle ABC holds R2r ≥ 1
8 + s2+r2
16Rr ≥ 1 (A new refinemento Euler’s R ≥ 2r inequality).
Mihaly Bencze
PP28332. In all triangle ABC holds�4R+rsr
�2 ≥ 2r
�2r − 1
R
�.
Mihaly Bencze
PP28333. If ai > 0 (i = 1, 2, ..., n) andP
cyclic
a1a2 = k then
Pcyclic
�a1 +
1a2
�2≥ (n+k)2
k .
Mihaly Bencze
PP28334. If x > 0 and λ ∈ (−∞, 0] ∪ [1,+∞) then
e−λx +�
272e2x+5ex+2
�λ≥ 23λ−1
�x
ex−1
�λ.
Mihaly Bencze
PP28335. If x ≥ 1 then 2x2 + 16x+ 1x + 1
x+2 + 12x+1 ≥ 5 5
√4 lnx.
Mihaly Bencze
PP28336. If a, b, c > 0 then2P
a2 3pb6 + 15b2c2 (b2 + c2) + c6 ≥P a4 + 3
Pa2b2.
Mihaly Bencze
Proposed Problems 847
PP28337. Prove thatnP
k=1
�8 ln(k2+k)
(k2+k−1)(k2+k+1)− 27
(k2+k+2)(2k2+2k+3)
�< n
n+1 .
Mihaly Bencze
PP28338. Prove thatn+1Pk=2
1k < n
q1− 1
n√2.
Mihaly Bencze
PP28339. If ak > 0 (k = 1, 2, ..., n) andnP
k=1
ak = n then
Pcyclic
an−11
a1+ n√a2·a3·...·an ≥ n
2 .
Mihaly Bencze
PP28340. If a, b > 0 then
3�a2 + b2
�≥ a
3
qa�2a+ 3
�√3− 1
�b�2
+ b3
qb�2b+ 3
�√3− 1
�a�2.
Mihaly Bencze
PP28341. Prove thatnP
k=1
k2 ln kk2+16k+1
≤ n(n−1)(2n+5)192 .
Mihaly Bencze
PP28342. Solve in (1,+∞) the following system:
x12 + 27
4 = 4 lnx2 +27
2(x3+2) +27
4(2x4+1) +1
2x5x22 + 27
4 = 4 lnx3 +27
2(x4+2) +27
4(2x5+1) +1
2x6
−−−−−−−−−−−−−−−−−−−xn
2 + 274 = 4 lnx1 +
272(x2+2) +
274(2x3+1) +
12x4
.
Mihaly Bencze
PP28343. If x ∈�0, π2
�then
sinx3 cos2 x−
√2 cosx+6
√2+ cosx
6+(6−√2) sinx
+√2
3 sin2 x−sinx cosx+6 cosx≥ 3
8 .
Mihaly Bencze
848 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP28344. If 1 ≤ a ≤ b thenb2−a2
4 − 272 ln b+2
a+2 − 278 ln 2b+1
2a+1 − 12 ln
ba + 43(b−a)
4 ≥ 4 ln bb
aa .
Mihaly Bencze
PP28345. If x ∈�0, π2
�then sin4 x
cos2 x+ cos4 x
2 + 4sin2 x
+ 2 sin2 x cos2 x ≥ 5.
Mihaly Bencze
PP28346. If x ∈�0, π4
�then cos 2x
ln(ctgx) ≥4(2+5 sinx cosx)1+16 sinx cosx .
Mihaly Bencze
PP28347. Denote M the set of all integers in {1, 2, ..., n} which arerelatively prime to n, and
M1 = M ∩�0, kn
�,M2 = M ∩
�kn ,
2kn
�, ...,Mn = M ∩
�(k−1)n
n , ni, where k ≥ 3.
Prove that if cardM ≡ 0 (mod k) then cardM1 = cardM2 = ... = cardMn.
Mihaly Bencze
PP28348. Solve in Z the equation
2�x2 + y2 + z2
�+ xy + yz + zx =
�x+y3 + 1
�3+�y+z
3 + 1�3
+�z+x3 + 1
�3.
Mihaly Bencze
PP28349. If x ∈�0, π2
�then
sin4 x
sin2 x+3√2 cos2 x
+ cos4 x
cos2 x+3√2 sin2 x
+ 4
2+3√sin2 x cos2 x
≥ 32 .
Mihaly Bencze
PP28350. In all triangle ABC holdsP a2
m2a+w2
a≤ 2
�Rr − 1
�.
Mihaly Bencze
PP28351. If a, b > 12 then�
a(a+1)2a−1
�2+
4(a2+b)(a+b2)(a+b−1)2
+�b(b+1)2b−1
�2≥ 6 (a+ b)− 3
2 .
Mihaly Bencze
PP28352. Determine all a, b, c ∈ N for which 10a+b+10b+c+10c+a
a2+b2+c2∈ N.
Mihaly Bencze
Proposed Problems 849
PP28353. If x ∈ R then 2�1 + sin2 x cos2 x
�≥ 3
3−sin2 x cos2 x.
Mihaly Bencze
PP28354. Let be ak > 0 (k = 1, 2, ..., n) and denote Fk the kth Fibonaccinumber. Prove that exist ik ∈ N (k = 1, 2, ..., n) such thatnP
k=1
(−1)Fik a2k ≥�
nPk=1
(−1)Fik ak
�2
.
Mihaly Bencze
PP28355. Determine all functions f : R → R for which
(x+ y + z) f (x+ y + z) = f�x2
�+f
�y2�+f
�z2�+2xf (y)+2yf (z)+2zf (x)
for all x, y, z ∈ R.
Mihaly Bencze
PP28356. If xk ∈�0, π2
�(k = 1, 2, ..., n) , then
nPk=1
(sinxk+cosxk)(1−sinxk cosxk)
sin2 xk cos2 xk≤
√n− 1
1
n�
k=1sin2 xk
+ 1n�
k=1cos2 xk
.
Mihaly Bencze
PP28357. If a, b, c > 0 then 2P
a2 (1− b)2 +P
(b− a)2 ≥ 13 (P |a− b|)2 .
Mihaly Bencze
PP28358. If a, b, c > 1 then(a−1)(b−1)(c−1)( 3√
abc−1)3
(√ab−1)
2(√bc−1)
2(√ca−1)
2 ≥ ln a ln b ln c(ln 3√abc)
3
ln2√ab ln2
√bc ln2
√ca.
Mihaly Bencze
PP28359. In all acute triangle ABC holds
1). 2P 1
sin A2
≤ (s−r)2−4R2
s2−(2R+r)2+ 2s
r + 3�2−
√3�
2). (s−r)2−4R2
s2−(2R+r)2≥ 3
�2−
√3�
Mihaly Bencze
850 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP28360. In all triangle ABC holdsP�
ma
wa
�λ≤ 3
�16r
q(R+2r)(s2+2Rr+5r2)
2R
�λ
for all λ ∈ [0, 1] .
Mihaly Bencze
PP28361. Let be x0 = 0, x1 = 1 and xn+1 = (xn + xn−1)qx2n + x2n−1 for all
n ≥ 1. Compute limn→0
xn.
Mihaly Bencze
PP28362. Solve in R the following system:
1
(x41+x4
2)(x1+x2)2 + 1
2(x22+x2
3)3 = 3
16x23x
24
1
(x42+x4
3)(x2+x3)2 + 1
2(x23+x2
4)3 = 3
16x24x
25
−−−−−−−−−−−−−−−−1
(x4n+x4
1)(xn+x1)2 + 1
2(x21+x2
x)3 = 3
16x22x
23
.
Mihaly Bencze
PP28363. In all triangle ABC holds
1).Pq
1(a4+b4)(c+b)2
+ 12(a2+b2)3
≤√3
8Rr
2).Pr
1
((s−a)4+(s−b)4)c2+ 1
2((s−a)2+(s−b)2)3 ≤
√3
4r2
3).Pr
1
(r4a+r4b)(ra+rb)
2 + 1
2(r2a+r2b)
3 ≤ (4R+r)√3
4s2r
4).Pr
1
(h4a+h4
b)(ha+hb)2 + 1
2(h2a+h2
b)3 ≤ (s2+r2+4Rr)
√3
16s2r2
Mihaly Bencze
PP28364. In all triangle ABC holds
1).P�
1(a4+b4)(a+b)2
+ 12(a2+b2)3
�≤ 3(s2−r2−4Rr)
128s2R2r2
2).P�
1
((s−a)4+(s−b)2)c2+ 1
2((s−a)2+(s−b)2)3
�≤ 3(s2−2r2−8Rr)
16s2r4
3).P�
1
(r4a+r4b)(ra+rb)
2 + 1
2(r2a+r2b)
3
�≤ 3((4R+r)2−2s2)
16s4r2
Mihaly Bencze
Proposed Problems 851
PP28365. Prove that there exist infinitely many positive integers n and ksuch that the largest prime divisor of n4 + n2k2 + k4 is equal to the largestprime divisor of (n+ k)4 + (n+ k)2 + 1.
Mihaly Bencze
PP28366. If x, y > 0 then 1(x4+y4)(x+y)2
+ 12(x2+y2)3
≤ 316x3y3
.
Mihaly Bencze
PP28367. In all triangle ABC holds
�2(s2−r2−Rr)s2+r2+2Rr
�λ
+�rR
�λ ≤ 2λ · 31−λ for
all λ ∈ [0, 1] .
Mihaly Bencze
PP28368. Determine all function f : R → R for which2f4 (x) + 4 = 3x3 + f
�x2
�+ 4f (x) for all x ∈ R.
Mihaly Bencze
PP28369. Determine the functions f : R → R such that1−x√
4f2(x)+45= 1
(x+1)2
qf(x)+15−3x for all x ∈ R\{5
3}.
Mihaly Bencze
PP28370. Let p ≥ 31 be a prime, andMk = {p+ k, 2p+ k, 3p+ k, ..., (p− 3) p+ k} . Determine all k ∈ N for whichone element of the set Mk is the sum of two integer squares.
Mihaly Bencze
PP28371. If x ∈ [0, 1] then 1−x√4x2+45
≥ 1(x+1)2
qx+15−3x .
Mihaly Bencze
PP28372. If x ∈ R thensin2 x√
45+4 cos2 x+ cos2 x√
45+4 sin2 x≥ 1
(1+sin2 x)2
q1+sin2 x2+3 cos2 x
+ 1(1+cos2 x)2
q1+cos2 x2+3 sin2 x
.
Mihaly Bencze
852 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP28373. If 0 ≤ a ≤ b ≤ 1 thenbRa
�q5−3xx+1 + 2
√4x+ 5
�dx ≥
12
�b√4b2 + 45− a
√4a2 + 45
�+ 45
4 ln�
2b+√4b2+45
2a+√2a2+45
�.
Mihaly Bencze
PP28374. Solve in R the following system:
q5−3x11+x1
+ 2√5 + 4x2 =
p4x23 + 45q
5−3x21+x2
+ 2√5 + 4x3 =
p4x24 + 45
−−−−−−−−−−−−−−−−q5−3xn
1+xn+ 2
√5 + 4x1 =
p4x22 + 45
.
Mihaly Bencze
PP28375. Prove thatnP
k=1
�q5n−3kn+k + 2
q5n+4k
n
�2
≥ 139n2+6n+23n .
Mihaly Bencze
PP28376. In all triangle ABC holds1). 27
�rs
�2+ 2 ≥ 1
3r3� 1
r3a
2). 27Rr2s2
+ 2 ≥ 13r3
� 1
h3a
Mihaly Bencze
PP28377. In all triangle ABC holds1). s2
�s2 + r (4R+ r)
�≥ 2r (4R+ r)2
2). 2 (sr)2�5s2 + r2 + 4Rr
�≥ r
�s2 + r2 + 4Rr
�2
Mihaly Bencze
PP28378. If a, b, c > 0 then 16 (P
a)2 + 18abc ≥ 9P
ab (a+ b) + 36P
ab.
Mihaly Bencze
PP28379. Determine all a, b ∈ N for which 5a
b3+b+1+ 5b
a3+a+1∈ N.
Mihaly Bencze
Proposed Problems 853
PP28380. In all triangle ABC holdsP
ma + 2P m2
a
b+c ≥ 4sr(s2−r2−Rr)R(s2+r2+2Rr)
.
Mihaly Bencze
PP28381. If a, b, c > 0 thenP a
3b2+2√3c√a2+b2+c2−bc
≥ 3√3
8√a2+b2+c2
.
Mihaly Bencze
PP28382. In all triangle ABC holds1).
Q �a2 + b2
�≥ 2s2
�s2 + r2 − 6Rr
�2
2).Q�
(s− a)2 + (s− b)2�≥ s2
�s2 + 20Rr
�2
3).Q �
r2a + r2b�≥ 16s4 (2R+ r)2
Mihaly Bencze
PP28383. Let ABCD be a square, and P ∈ BC, Q ∈ CD such thatPAQ∡ = x. Let E and F be the intersections of PQ with AB and AD,respectively. Determine all x such that (AE +AF )2 ≥ 8Area [ABCD] .
Mihaly Bencze
PP28384. In all triangle ABC holds
2Q �
cos2 A2 + cos2 B
2
�≥
�Pcos3 A
2 +Q �
cos A2 + cos B
2 − cos C2
��2.
Mihaly Bencze
PP28385. In all triangle ABC holds2Q �
m2a +m2
b
�≥
�Pm3
a +Q
(ma +mb −mc)�2
.
Mihaly Bencze
PP28386. If a, b, c > 0 then 2Q �
a2 + b2�≥ (P
ab (a+ b)− 2abc)2 .
Mihaly Bencze
PP28387. If a, b, c > 0 then�P
a2b− abc�2
+�P
ab2 − abc�2 ≤ 8
27
�Pa2�3
.
Mihaly Bencze
PP28388. Find all functions f : N → N such that m2 + f (mn) + n2 isdivisible by mf (m) +mn+ nf (n) for all m,n ∈ N ∗.
Mihaly Bencze
854 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP28389. If a, b, c > 0 then�P
a2b− abc�2
+�P
ab2 − abc�2 ≥ 8a2b2c2.
Mihaly Bencze
PP28390. Solve in Z the equation
2P
x2 +P
xy =�x+y
2 + 1�3
+�y+z
2 + 1�3
+�z+x2 + 1
�3.
Mihaly Bencze
PP28391. If x, y, z ≥ 14 then
P ((x+y)2+y+z)((y+z)2+x+y)(x+2y+z−1)2
≥ 4P
x− 34 .
Mihaly Bencze
PP28392. If x, y, z ≥ 13 then
P 1√y−z+(x+y+z)z
≤qP 1
xy .
Mihaly Bencze
PP28393. If x, y, z ≥ 13 then
Pqz
(y−z+(x+y+z)z)(z−x+(x+y+z)) ≤1√xyz .
Mihaly Bencze
PP28394. If x, y, z ≥ 13 thenP
(yP
x+ z − y) (zP
x+ y − z) ≥ (P
xy) (P
x)2 .
Mihaly Bencze
PP28395. If x, y, z ≥ 13 then
P yzy−z+z(x+y+z) ≤ 1.
Mihaly Bencze
PP28396. If x, y, z ≥ 13 then
P z(y+(x+y+z−1)z)(z+(x+y+z−1)x) ≤
(�
x−1)�
x2+((�
x)2−�
x+1)�
xy
xyz(�
x)4.
Mihaly Bencze
PP28397. Prove that
�nQ
k=1
kk�4
≤ exp
�nP
k=1
(k2−1)(k2+16k+1)(k+2)(2k+1)
�.
Mihaly Bencze
Proposed Problems 855
PP28398. If 1 ≤ a ≤ b thenb2−a2
4 − 272 ln b+2
a+2 − 278 ln 2b+1
2a+1 − 12 ln
ba + 27
4 (b− a) ≥ 4 ln�
eabb
ebaa
�.
Mihaly Bencze
PP28399. Prove thatnP
k=1
�8 ln k(k+1)
(k2+k−1)(k2+k+1)− 27
2k4+4k3+7k2+5k+2
�≤ n
n+1 .
Mihaly Bencze
PP28400. Prove that (n!)4 ≤ exp
�nP
k=1
(k2−1)(k2+16k+1)k(2k2+5k+2)
�.
Mihaly Bencze
PP28401. If xk ∈ (0, 1) (k = 1, 2, ..., n) , then
1).nP
k=1
(1 + xk)2 lnxk ≤ 2
�nP
k=1
x2k − n
�
2).nP
k=1
(1 + xk)�1 + xk + x2k
�lnxk ≤ 2
�nP
k=1
x3k − n
�
Mihaly Bencze
PP28402. If xk ≥ 1 (k = 1, 2, ..., n) thennQ
k=1
xk ≥ exp
2
�
n�
k=1xk−n
�2
n�
k=1x2k−n
.
Mihaly Bencze
PP28403. If xk ∈ (0, 1) (k = 1, 2, ..., n) , thennP
k=1
(1 + xk) lnxk ≤ 2
�nP
k=1
xk − n
�.
Mihaly Bencze
PP28404. Prove thatnP
k=1
�1 + 1
k(k+1)
�ln 1
k(k+1) ≤−2n2
n+1 .
Mihaly Bencze
PP28405. In all triangle ABC holdsP
(π +A) ln Aπ < −4π.
Mihaly Bencze
856 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP28406. In all triangle ABC holds
1).P �
1 + sin2 A2
�ln�sin A
2
�< −2− r
2R
2).P �
1 + cos2 A2
�ln�cos A
2
�< −1 + r
2R
Mihaly Bencze
PP28407. If ak > 0 (k = 1, 2, ..., n) then
Pcyclic
aλ+11
(a1+ n√a2a3...an)
λ ≥nλ
�
n�
k=1ak
�λ+1
�
n+(2n−1)n�
k=1ak
�λ for all λ > 0.
Mihaly Bencze
PP28408. In all triangle ABC holds
1).P�
rra
+ 1�ln r
ra< −4
2).P�
rha
+ 1�ln r
ha< −4
Mihaly Bencze
PP28409. Determine all functions f : N ∗ → N∗ such that(n− 1)3 < f (n) (f ◦ f) (n) ((f ◦ f ◦ f)n) < n3 + n2 + n+ 1 for all n ∈ N∗.
Mihaly Bencze
PP28410. If λ ∈ [0, 1] thennP
k=1
1k2λ
≤ n�1− 1
n√2
�λ.
Mihaly Bencze
PP28411. Let ABC be a triangle in which B∡ = xC∡, CD = yBC whenextend the side BC be a segment CD. Determine all x, y ∈ R for whichtgBAD∡
2 = a2
4sr .
Mihaly Bencze
PP28412. Determine all functions f : N → N for which
f
nP
k=1
a2k +P
1≤i<j≤naiaj
!=
nPk=1
f2 (ak) + f
P
1≤i<j≤naiaj
!for all
ak ∈ N (k = 1, 2, ..., n) .
Mihaly Bencze
Proposed Problems 857
PP28413. Solve in Z the equation (x+yz)2
x+z + (y+zx)2
y+x + (z+xy)2
z+y = x2 + y2 + z2.
Mihaly Bencze
PP28414. Solve in Z the equation
x�y+z
2
�2+ y
�z+x2
�2+ z
�x+y2
�2= x3 + y3 + z3.
Mihaly Bencze
PP28415. If a, b > 0 then ab(a+b)2
+ 2√ab
(√a+
√b)
2 ≤ 14 + 4ab
(a+b)(√a+
√b)
2 .
Mihaly Bencze
PP28416. In all triangle ABC holds
4P �
ab + 1 + b
a
�3 ≥ 54P a2+b2
ab ≥ 27(s2+r2+10Rr)2Rr .
Mihaly Bencze
PP28417. If x0 = 1 and xkn+1 + 1 = (xn + 1)k for all n ≥ 0 then determineall k ∈ R for which [xn] = n for all n ≥ 1, where [·] denote the integer part
Mihaly Bencze and Tibor Jakab
PP28418. Solve in Z the equationnP
k=1
x2k = n− 2 +P
1≤i<j≤nxixj .
Mihaly Bencze
PP28419. If ak ∈ [0, 1] (k = 1, 2, ..., n), then
P
1≤i<j≤n
11+aiaj
! P
1≤i<j<k≤n
11+aiajak
!≥ n2(n−1)2(n−2)
72 .
Mihaly Bencze
PP28420. Solve in Z the equationP
cyclic
x1x2...xn−1 = n− 1 +nQ
k=1
xk.
Mihaly Bencze
858 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP28421. If a, b > 0 then��√3− 1
�a+
�2√3 + 1
�b�q
aa+2b +
��√3− 1
�b+
�2√3 + 1
�a�q
bb+2a ≤
3 (a+ b) .
Mihaly Bencze
PP28422. If a, b, c > 0 then a3(6c−a)2
(8c2+a2)2+ b3(6a−b)2
(8a2+b2)2+ c3(6b−c)2
(8b2+c2)2< a+ b+ c.
Mihaly Bencze
PP28423. If a, b, c > 0 and a+ b+ c = 2 then 27�a2 + b2 + c2
�+64abc ≥ 54.
Mihaly Bencze
PP28424. Compute limt→0
1t
�3t
tR0
√1 + exdx+ 5
t
tR0
eexdx− 3
√2− 5e
�.
Mihaly Bencze
PP28425. If xk > 0 (k = 1, 2, ..., n) , thennP
k=1
x2k√
1+xk≥
�
n�
k=1xk
�2
�
n2+nn�
k=1xk
.
Mihaly Bencze
PP28426. If x, y, z > 0 thenP x2√
1+x+ (
�
x)2√9+3
�
x≥P (x+y)2√
4+2(x+y).
Mihaly Bencze
PP28427. If ai > 0 (i = 1, 2, ..., n) and k ∈ N, k ≥ 2,nP
i=1ai = 1 then
Qcyclic
�1 + ak2a
k3...a
kn
�a1 ≥ 1 +
n
nQp=1
ap
!k
.
Mihaly Bencze
PP28428. If x, y ∈ (0, π2 ) then:
(sin2 x+sin2 y)sin2 + sin2 y ·(cos2 x+cos2 y)cos
2 x+cos2 y
(sinx)2 sin2 x·(sin y)2 sin2 y ·(cosx)2 cos2 x·(cos y)2 cos2 y≤ 4
Daniel Sitaru
Proposed Problems 859
PP28429. If a, b, c > 0; a+ b+ c = 3; 0 ≤ x ≤ 1 then:a�ba
�x+ b
�cb
�x+ c
�ac
�x+ b
�ab
�x+ c
�bc
�x+ a
�ca
�x ≤ 6
Daniel Sitaru
PP28430. Find: Ω =P∞
k=1
�P∞n=1n6=k
�1
n2−k2
��
Daniel Sitaru
PP28431. Find: Ω =R �P∞
n=1
�3n sinh3
�x3n
���dx
Daniel Sitaru
PP28432. If x, y, z, t > 0 then:
(xy + yz + zt+ tx)�
1x4 + 1
y4+ 1
z4+ 1
t4
�≥
�1x + 1
y + 1z + 1
t
�2
Daniel Sitaru
PP28433. If a, b, c, d > 0 then:(ab+ bc+ cd+ da)(a4 + b4 + c4 + d4) ≥ abcd(a+ b+ c+ d)2
Daniel Sitaru
PP28434. Let be: Ω(a, b, c) =P∞
n=1an2+bn+c
n! ; a, b, c > 0
Prove that: Ω(a, b, c) + Ω(b, c, a) + Ω(c, a, b) ≥ 3(4e− 1) 3√abc
Daniel Sitaru
PP28435. In ΔABC the following relationship holds:P
CyC(a,b,c) aq�
(b− c)2 + 4r2��(c− a)2 + 4r2
�≥ abc
Daniel Sitaru
PP28436. In acute ΔABC the following relationship holds:2sinA + ssinB + 2sinC + 2cosA + 2cosB + 2cosC > 9
Daniel Sitaru
PP28437. If a1, a2, ..., a8 ≥ 1 then: a41 + a42 + ...+ a48 ≤ (a1a2...a8)4 + 7
Daniel Sitaru
860 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP28438. Find: Ω = limn→∞ npΩn(a)− (a+ 1)!; a ∈ N;
Ωn(a) =Pn
k=0(k2 − a2 + 1)(a+ k)!; n ∈ N
Daniel Sitaru
PP28439. Let be Ω(a) = limn→∞Pn
k=1
�1
3k sin3(3k sin a)
�
Prove that if a, b, c ∈ [0, π2 ] then: 4�bΩ(a) + cΩ(b) + aΩ(c)
�≤ 3(a2 + b2 + c2)
Daniel Sitaru
PP28440. If 0 < a ≤ b ≤ c then: 1(a−b+c)6
+ 1b6
≤ 1a6
+ 1c6
Daniel Sitaru
PP28441. If a, b > 0, a2 + b2 = 1 then: 1a + 1
b ≥ 2√2.
Daniel Sitaru
PP28442. If x, y > 0; Ω(x, y) =P∞
n=12n2+(2x+2y+5)n+2xy+6x−y3n(n+y)(n+y+1)(n+y+2) then:
Ω(y, x) ≤ 19 3√xy
Daniel Sitaru
PP28443. If a, b, c, d > 0; a3 + b3 + d3 = 1 then: a+b+c+dabcd ≥ 16
Daniel Sitaru
PP28444. If a, b, c, > 0; a3 + b3 + c3 = 1 then: a+b+cabc ≥ 3 3
√9
Daniel Sitaru
PP28445. If a, b, c, d > 0; a2 + b2 + c2 + d2 = 1 then: a+b+c+dabcd ≥ 32
Daniel Sitaru
PP28446. If a, b, c > 0; a2 + b2 + c2 = 1 then: a+b+cabc ≥ 9
Daniel Sitaru
PP28447. If a, b, c > 0; x, y, z, t ∈ R then:(a+b+c)y2
a + (a+b+c)z2
b + (a+b+c)t2
c ≥ x(2y + 2z + 2t− x)
Daniel Sitaru
Proposed Problems 861
PP28448. If a, b > 0; x, y, z ∈ R then: (a+b)y2
a + (a+b)z2
b ≥ x(2y + 2z − x)
Daniel Sitaru
PP28449. In ΔABC the following relationship holds:
3
qsinAsinB + 3
qsinBsinC + 3
qsinCsinA − 3
qsinAsinC − 3
qsinBsinA − 3
qsinCsinB < 1
Daniel Sitaru
PP28450. If x ∈�0,√3�then�
1(1+x)2
+ 2x2
(3+2x−x2)
��1
(1+x)2+ 8x2
(3+2x−x2)2
�≥ 9
32 .
Mihaly Bencze
PP28451. If a, b > 0 and 2ab+ a2 = 2a+ b then�2
1+a + 11+b
��2
1+b +1
1+a
�≤ 25
9 .
Mihaly Bencze
PP28452. Prove thatnP
k=1
sin π2k(k+1) >
nn+1 .
Mihaly Bencze
PP28453. If xk > 0 (k = 1, 2, ..., n) andnP
k=1
xk ≤ 1 then
nQk=1
�1− 1
xλk
�≥
�2λ − 1
�nfor all λ ≥ 1.
Mihaly Bencze
PP28454. If m,n, k ∈ N ;m,n, k ≥ 2 then�m3 − 1
� �n3 − 1
� �k3 − 1
�≥ 5m2n2k2 + 2mnk − 1.
Mihaly Bencze
PP28455. If m,n, k ∈ N ;m,n, k ≥ 2 then�mk − 1
� �nk − 1
�≥ (2k−1−1)mk−1nk−1
2k−3 + 1.
Mihaly Bencze
862 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP28456. In all convex quadrilateral ABCD holds1 + 2
PsinA+
Psin2A ≤ 3
Q(1 + sinA) .
Mihaly Bencze
PP28457. Solve in C the following system:
x+ y2 + z3 = 32x2 + y3 + z4 = 74x3 + y4 + z5 = 260
.
Mihaly Bencze
PP28458. If a, b > 0 and ab = 2 then 4a+b +
a+b2ab ≤ 3(a+b)
2 .
Mihaly Bencze
PP28459. Denote Fk and Lk the kth Fibonacci respective Lucas numbers.Prove that:
1).qx2 + xFk + F 2
k +qx2 + xFk+1 + F 2
k+1 ≥q
4x2 + 2xFk+2 + F 2k+2
2).qx2 + xLk + L2
k +qx2 + xLk+1 + L2
k+1 ≥q4x2 + 2xLk+2 + L2
k+2 for
all x ≥ 0
Mihaly Bencze
PP28460. For a positive integer n, we define Fa (n) to be the number of a′ sthat appear as digits after writting the numbers 1, 2, ..., n in their decimalexpansion. Prove that there are finitely many numbers m such thatFa (m) = m for all a a ∈ {2, 3, 4, 5, 6, 7, 8, 9} .
Mihaly Bencze
PP28461. If a0 = 3, a1 = a, a2 = b and an = aan−1 + ban−2 + can−3 for alln ≥ 3 then write an as a polynomial in a, b, c.
Mihaly Bencze
PP28462. In all triangle ABC holds6√14 ≤P
√15 + cos 4A+
P√15− cos 4A ≤ 24.
Mihaly Bencze
Proposed Problems 863
PP28463. If a, b, c > 0 then�5−
√5�P
a2 +√5min
�a2, b2, c2
≥ 3
�√5− 1
�Pab.
Mihaly Bencze
PP28464. If a, b, c > 0 then√a2 + ab+ b2 +
√b2 + bc+ c2 ≥
q(a+ c)2 + 2 (a+ c) b+ 4b2.
Mihaly Bencze
PP28465. Denote Fk and Lk the kth Fibonacci respectve Lucas numbers.
Prove thatqx2 + xFk + F 2
k +qx2 + xLk + L2
k +q
x2 + xFk+1 + F 2k+1+
+qx2 + xLk+1 + L2
k+1 ≥q
16x2 + 4x (Fk+2 + Lk+2) + (Fk+2 + Lk+2)2,
for all x ≥ 0.
Mihaly Bencze
PP28466. In all triangle ABC holdsP (ha−ra)(ha+hara+ra)
hara�
ha
√r2a+ra+1+ra
√h2a+ha+1
� ≥ 0.
Mihaly Bencze
PP28467. In all triangle ABC holdsQ
cyclic
�ra
qr2b + rb + 1 + rb
pr2a + ra + 1
�≥ 4Rs2
r Qcyclic
(h2a + ha + 1).
Mihaly Bencze
PP28468. Prove that if a, b, c > 1, thenP
loga c ≥P
logab b2. Where
n ∈ N ∗ [1] , a ∈ R\ [0, 1] .
Onose Alina-Iulia
PP28469. Let x, y ∈�0, π2
�. Prove that:
1cosx(sinx+cosx) +
1cos y(sin y+cos y) ≥ 4 tan π
8 .
Onose Alina-Iulia
PP28470. If a, b, c > 0 and a+ b+ c = 27, prove that:ab +
bc +
ca + 3 ≥ 2
3
�√a+
√b+
√c�.
Onose Alina-Iulia
864 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP28471. If A,B,C are the angles of a triangle, show that the relationshipbelow is true: (cosA+ cosB − cosC)2 + (cosA− cosB + cosC)2+
+(− cosA+ cosB + cosC)2 ≥ 14 + 4 sin A
2 sin B2 sin C
2 .
Maria Mantu and Ioana Mantu
PP28472. Let n be an integer with n ≥ 3 and let ai, i = 1, 2, ..., n, bepositive real numbers such that a1 ≤ a2 ≤ ... ≤ an. Prove that for any realnumber k with k > 1 the following inequality holds:
a1ak2 + a2a
k3 + ...+ an−1a
kn + ana
k1 ≥ ak1a2 + ak2a3 + ...+ akn−1an + akna1
Leonard Giugiuc
PP28473. Let a, b and c be real numbers situated in the interval [1,∞) suchthat a+ b+ c = 1 + 1
a + 1b +
1c . Find the maximum value of the expression
ab+ bc+ ca.
Sladjan Stankovik, Leonard Giugiuc and Diana Trailescu
PP28474. Prove that if k =�√
7 +√3− 4
� �2 +
√3�, then√
x2 + 3 +py2 + 3 + k
√xy + 3 ≥ 2k + 4 for all non negative real numbers x
and y with x+ y = 2. When equality holds?
Leonard Giugiuc, Marius Dragan and Marian Cucoanes
PP28475. In all acute triangle ABC holdsP A(3−A)B+C + 9−π
2 ≥P (A+B)(6−A−B)π+C .
Mihaly Bencze
PP28476. If a, b, c > 0 and λ > 0 thenP a√
a2+λbc+ (P
a)q
�
a�
a3+3λabc≥P (a+b)
√a+b√
a3+b3+2λabc.
Mihaly Bencze
PP28477. In all acute triangle ABC holds3√3
2π + 12
P sinAB+C ≤ 4
3
P cos A2
B+C ≤ 9√3
2π .
Mihaly Bencze
Proposed Problems 865
PP28478. In all triangle ABC holdsP
a2x ≥ 4srqP 5−3yz
1+x for all
x, y, z > 0 and x+ y + z = 3.
Mihaly Bencze
PP28479. If ak, bk, ck > 0 (k = 1, 2, ..., n) , then
nPk=1
akbkck + 3
s�nP
k=1
a3k
��nP
k=1
b3k
��nP
k=1
c3k
�≥
≥ 2
n2
nX
k=1
ak
! nX
k=1
bk
! nX
k=1
ck
!.
Mihaly Bencze
PP28480. In all triangle ABC holdsP
a2x ≥ 4sr 10p12P
x4y4z (x+ y), forall x, y, z > 0.
Mihaly Bencze
PP28481. In all triangle ABC holdsP
a2x ≥ 4sr
rP xy
√xy(xy+3z2)
(x+y)z for all
x, y, z > 0.
Mihaly Bencze
PP28482. In all triangle ABC holdsP
a2x ≥ 4sr 8
q818
Q(1 + x2y2) for all
x, y, z > 0 and xyz = 1.
Mihaly Bencze
PP28483. In all triangle ABC holdsP
a2x ≥ 2srq
3√3�
(x+yz)2�
xy for allx, y, z > 0 and x+ y + z = 3.
Mihaly Bencze
PP28484. In all triangle ABC holdsP
a2x ≥ 8srq
15
P x√y√
x+zfor all
x, y, z > 0.
Mihaly Bencze
866 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP28485. In all triangle ABC holdsP
a2x ≥ 2srq5P x(y+z)
x+1 − 3xyz whenx, y, z > 0 and x+ y + z = 3.
Mihaly Bencze
PP28486. If a, b, c, d > 0 then3�a2 − ab+ b2
�2+ 2abcd+ 3
�c2 − cd+ d2
�2 ≥ 2�a2c2 + b2d2
�.
Mihaly Bencze
PP28487. In all triangle ABC holdsP
a2x ≥ 4srq
12
P x2(y+z)2
y2+z2for all
x, y, z > 0.
Mihaly Bencze
PP28488. In all triangle ABC holdsP
a2x ≥ 4srpP
xλ when x, y, z > 0,x+ y + z = 3 and λ = 2 ln 3−3 ln 2
ln 3−ln 2 .
Mihaly Bencze
PP28489. In all triangle ABC holdsP
a2x ≥ 12sr
√�
x2� x
y
for all x, y, z > 0.
Mihaly Bencze
PP28490. In all triangle ABC holds
Pa2x ≥ 8sr
p2 (P
x2y2) (P
x3 + 16xyz) for all x, y, z > 0 and x+ y+ z = 1.
Mihaly Bencze
PP28491. In all triangle ABC holdsPa2x ≥ 4 4√3sr√
26p(x+ y) (y + z) (z + x) for all x, y, z > 0.
Mihaly Bencze
PP28492. In all triangle ABC holdsP
a2x ≥ 4sr
r3√
�
(x+y)�
x2(y+z)for all
x, y, z > 0.
Mihaly Bencze
Proposed Problems 867
PP28493. Determine all triangle ABC in whichPa2x ≥ sr
p2 (P
x3 + 9xyz +P
x) for all x, y, z > 0.
Mihaly Bencze
PP28494. In all triangle ABC holdsPa2py2 − yz + z2 ≥ 4sr
px2 + y2 + z2 for all x, y, z > 0.
Mihaly Bencze
PP28495. In all triangle ABC holdsP
a2py2 + yz + z2 ≥ 4sr
Px for all
x, y, z > 0.
Mihaly Bencze
PP28496. In all triangle ABC holdsP
a2x ≥ 4srq
�
a2b2
(�
a2y)(�
a2z)for all
x, y, z > 0.
Mihaly Bencze
PP28497. In all triangle ABC holdsP
a2 ≥ 4srqabc+ 3
pQ(1 + a3).
Mihaly Bencze
PP28498. In all triangle ABC holdsP
a2x ≥ 4sr
rq23
P x√y√
x+27for all
x, y, z > 0.
Mihaly Bencze
PP28499. If a, b, c > 0 and a+ b+ c = 1 thenP �
a2 + b2� �
b2 + c2�≥ 24
�Pa2b2
� 43 .
Mihaly Bencze
PP28500. If a, b, c > 0 and abc = 1 thenP(a+ b)3 (b+ c)3 ≥ 18 (a+ b+ c− 1)2 .
Mihaly Bencze
PP28501. In all triangle ABC holdsP(x+ y) a2 ≥ 4
√3sr 6p4 (x+ y + z − 1) where x, y, z > 0 and xyz = 1.
Mihaly Bencze
868 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP28502. In all triangle ABC holdsP
a3 ≥ 4srq8s2Rr + 1
2
Pa2 (b− c)2.
Mihaly Bencze
PP28503. If a, b, c > 0 and abc = 1 thenP
(a+ b)3 ≥ 12 (a+ b+ c− 1) .
Mihaly Bencze
PP28504. In all triangle ABC holds (P
a cosA)2 ≤P
m2a.
Mihaly Bencze
PP28505. In all acute triangle ABC holdsP
cos 2A+ 2 (P
cosA)2 ≤ 3.
Mihaly Bencze
PP28506. In all triangle ABC holdsP�
m2a
bc + a2
mbmc
�≥ 25
4 .
Mihaly Bencze
PP28507. In all triangle ABC holdsP �r2a + r2b + r2c
�a �m2
a +m2b +mc
�b ≥ 3s8s3 .
Mihaly Bencze
PP28508. Let ABCD be a convex quadrilateral in which A∡ = C∡ = 90◦.Denote the proiection of A and C to BD with A1 and C1. Prove that
AA1 + CC1 +�2√2− 1
�BD ≤ Perimeter (ABCD) .
Mihaly Bencze
PP28509. In all triangle ABC holdsP a2(a+c)(a+c−b)
b ≥ 24sr
rsr(s2+r2+2Rr)� (c+a)(c+a−b)
b
.
Mihaly Bencze
PP28510. Solve in Z the equation x3+yzy2+z2
+ y3+zxz2+x2 + z3+xy
x2+y2= 0.
Mihaly Bencze
Proposed Problems 869
PP28511. In all triangle ABC holds
a2xz (x+ z)+b2yx (y + x)+c2zy (z + y) ≥ 2srq2 (P
xz (x+ z))pQ
(x+ y)for all x, y, z > 0.
Mihaly Bencze
PP28512. In all triangle ABC holds
a2 + 2n+16 b2 + n(n+1)
2 c2 ≥ 8srn(n+1)
nPk=1
pk3 (k2 + k + 1).
Mihaly Bencze
PP28513. In all triangle ABC holds
a2
n+1 + b2
m+1 + c2
k+1 ≥ 4srnmk
nPp=1
mPr=1
kPt=1
q1
pr(p+1)(r+1) +1
rt(r+1)(t+1) +1
pt(p+1)(t+1) .
Mihaly Bencze
PP28514. If A,B ∈ Mn (R) and AB = BA then detrmine all λ ∈ R for
which detnQ
k=0
�A2 + (λ+ k)AB + (λ+ k + 1)B2
�≥ 0.
Mihaly Bencze
PP28515. If A,B ∈ Mn (R) and AB = BA thendet
��A2 + 3AB + 4B2
� �A2 + 4AB + 5B2
��≥ 0.
Mihaly Bencze
PP28516. Let ABC be a triangle in which A ≤ B ≤ C. Prove that
sinB + sinC > (sinA)sinA(1−sinB)sinB(1−sinA) + (sinB)
sinB(1−sinC)sinC(1−sinB) .
Mihaly Bencze
PP28517. In all acute triangle ABC holdsPlog2
�2sinA + 2cosA
�≥ 3 + 1√
2
Psin
�A+ π
4
�.
Mihaly Bencze
PP28518. Prove thatnP
k=1
�k4 + 10k3 + 32k2 + 40k + 20
�((k + 1)!)2 =
�n2 + 6n+ 5
�((n+ 2)!)2−20.
Mihaly Bencze
870 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP28519. If ak > 0 (k = 1, 2, ..., n) andnQ
k=1
ak = 1 thenP
cyclic
a1ea2 ≥ ne.
Mihaly Bencze
PP28520. If a, b, c > 0 then
�a+ba
�2a � b+cb
�2b � c+ac
�2a ≤�2a+b+ca+c
�a+c �2b+c+ab+a
�b+a �2c+a+bc+b
�c+b.
Mihaly Bencze
PP28521. If xk ∈ R (k = 1, 2, ..., n) and xi 6= xj (i 6= j) then
Pcyclic
|x1 + x2| ≤P
cyclic
x1|x1|−x2|x2|x1−x2
≤ 2nP
k=1
|xk| .
Mihaly Bencze
PP28522. In all triangle ABC holdsP 1
sinA ≤ 2 + 1sinA sinB sinC .
Mihaly Bencze
PP28523. In all triangle ABC holds s2−8Rrr2
+�4R+r
s
�2 ≤ 2+16R2�
1s2
+ 1r2
�.
Mihaly Bencze
PP28524. If 0 ≤ ak ≤ bk (k = 1, 2, ..., n) thennP
k=1
ebk−eakbk−ak
≤ n− 1 +nQ
k=1
ebk−eakbk−ak
.
Mihaly Bencze
PP28525. If a, b, c > 0 and a+ b+ c = 1 thenP 1
a+1 ≤ 3918 + 4
3
P 14−a2
.
Mihaly Bencze
PP28526. If 1 ≤ ak ≤ bk (k = 1, 2, ..., n) then
12
nPk=1
(ak + bk) ≤ n− 1 + 12n
nQk=1
(ak + bk) .
Mihaly Bencze
Proposed Problems 871
PP28527. In all triangle ABC holds:
1). 3P ra
r+ra≤ 13
2 + 4minn
r2a4r2a−r2
;r2b
4r2b−r2
; r2c4r2c−r2
o
2). 3P ha
r+ha≤ 13
2 + 4minn
h2a
4h2a−r2
;h2b
4h2b−r2
; h2c
4h2c−r2
o
Mihaly Bencze
PP28528. If x ∈�0, π2
�then 17 + 112 sin2 x cos2 x ≥ 54
√2 sinx cosx.
Mihaly Bencze
PP28529. In all triangle ABC holds1). 27 + 3
�rs
�2+ 6r(4R+r)
s2≥ r3
P 1r3a
+ 54√2 rs
2). 27 +3(s2+r2+5Rr)
2s2≥ r3
P 1h3a+ 54
s
√Rr
Mihaly Bencze
PP28530. If 0 < x ≤ y < π2 then�q
xy + 1
� �sin y − sin
√xy
�≥ sin y − sinx ≥
�qyx + 1
� �sin
√xy − sinx
�.
Mihaly Bencze
PP28531. If x > 0 thenx6+x3+1
x3 + x2+x+1x + x4+x2+1
x2 ≥ 9�
x2
1+2x2 + 12+x + x
x3+x+1
�.
Mihaly Bencze
PP28532. If a, b, c ∈ [0, 1] then(1+a3)(1+b6)(1+c9)(1−a3)(1−b6)(1−c9)
≥ (1+ab2c3)(1+bc2a3)(1+ca2b3)(1−ab2c3)(1−bc2a3)(1−ca2b3)
.
Mihaly Bencze
PP28533. In all triangle ABC holds1). 324r4
P 1h4a+ 45 ≥ 171r2
P 1h2a
2). 324r4P 1
r4a+ 45 ≥ 171r2
P 1r2a
Mihaly Bencze
PP28534. If ak > 0 (k = 1, 2, ..., n) thenP
cyclic
4√a2a3
2a1+a2+a3≥ 1
4
nPk=1
1√ak.
Mihaly Bencze
872 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP28535. Prove that exp
�nP
k=1
�√k2 + k +
√k2 + 2k +
√k2 + 3k + 2
��≤
≤ en(3n+5)
2
2n
�(e− 1)3 (e+ 1)
�n
for all n ∈ N∗.
Mihaly Bencze
PP28536. Denote Fk and Lk the kth Fibonacci, respective Lucas numbers.Prove that
1). Fn+2 − 1 ≤nP
k=1
Fλ+1k
+Fλ+1k+1
Fλk+Fλ
k+1
≤ Fn+3 − 2
2). Ln+2 − 3 ≤nP
k=1
Lλ+1k
+Lλ+1k+1
Lλk+Lλ
k+1
≤ Ln+3 − 4 for all λ ≥ 1
Mihaly Bencze
PP28537. In all triangle ABC holdsP
ln2�1 +
√sinA
�≤ s
R .
Mihaly Bencze
PP28538. If f : R → (0,+∞) when 0 < a ≤ b, is continuous and increasing,
then (a+ 3b)abRaf (x) dx
4(a+b)2b2R
(a+b)2f (x) dx ≤ a
bRaf (x) dx
4b2R
(a+b)2f (x) dx.
Mihaly Bencze
PP28539. Solve in R the equation: (arcsin [x])3 (arccos [x])5 = x3�π2 − x
�5where [·] denote the integer part.
Mihaly Bencze and Rovsen Pirguliev
PP28540. In all triangle ABC holdsP 1
sin3 A2
+ 12Rr
P sin A2
sin B2+sin C
2
+ 24(R+r)r ≥ 216.
Mihaly Bencze
PP28541. In all triangle ABC holds 4s2r ≤ 3R�s2 − r (4R+ r)
�.
Mihaly Bencze
Proposed Problems 873
PP28542. In all triangle ABC holds s2λ + 5r2λ + 2 (Rr)λ ≤ 8�RrP m2
a
bc
�λ
for all λ ∈ [0, 1] .
Mihaly Bencze
PP28543. In all triangle ABC holds�P 1sin A
2
�3
≥ 24(R+r)r + 12R
r
P sin A2
sin B2+sin C
2
.
Mihaly Bencze
PP28544. In all triangle ABC holds
1).P
b sinA ≤√32
�(s2+r2+4Rr)
2
4sRr − 4s
�
2).P
(s− a) sinA ≤√32
�(4R+r)2−2s2
s
�
3).P
ra sinA ≤√32
�s2−2r(4R+r)
r
�
4).P
sinA sin2 A2 ≤
√32
��s2+r2−8Rr
4Rr
�2+ r
R − 2
�
5).P
cosA cos2 A2 ≤
√32
�(s2+(4R+r)2)
2−16Rs2(4R+r)
16s2R2
�
Mihaly Bencze
PP28545. In all acute triangle ABC holds9−π6 +
P sinAB+C ≥ 1
3
P (A+B)(6−A−B)π+C .
Mihaly Bencze
PP28546. In all triangle ABC holds 2P�
ma
mb
�2≥
�s2 − r2 − 4Rr
�P 1m2
a.
Mihaly Bencze
PP28547. In all acute triangle ABC holdsP
ctgA ≥ (�
cosA)2
2 sinA sinB sinC .
Mihaly Bencze
PP28548. In all acute triangle ABC holdsP
cos2A+ (P
cosA)2 ≤ 3.
Mihaly Bencze
874 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP28549. In all acute triangle ABC holds�Psin3A
� �Pcos3A
�2 ≥ (sinA sinB sinC)3 .
Mihaly Bencze
PP28550. In all acute triangle ABC holds (P
sinA)2 ≤ 2Q sin2 A
cosA .
Mihaly Bencze
PP28551. In all acute triangle ABC holds�2R2
sr
�λ+ (Q
ctgA)λ ≤ 21−λ (P
ctgA)λ for all λ ∈ [0, 1] .
Mihaly Bencze
PP28552. In all acute triangle ABC holds�Psin3A
� �P 1cos3 A
�2 ≥ 8tg3Atg3Btg3C.
Mihaly Bencze
PP28553. In all acute triangle ABC holds�P
cos3A� �P 1
sin3 A
�2≥ 8.
Mihaly Bencze
PP28554. In all acute triangle ABC holds8�P
sin3A�2 �P
cos3A�≥
�Psin2A
�3.
Mihaly Bencze
PP28555. Determine all λ1,λ2 > 0 for which in all acute triangle ABCholds 2
P(sinA)λ1 ≤ 3
P(ctgA)λ2 .
Mihaly Bencze
PP28556. In all triangle ABC holds r2λ +�s2−6Rr−3r2
4
�λ≤ 21−λR2λ for all
λ ∈ [0, 1] .
Mihaly Bencze
PP28557. In all triangle ABC holdsP
(ab)2λ ≤ 31−λ (RP
a)2λ for allλ ∈ [0, 1] .
Mihaly Bencze
Proposed Problems 875
PP28558. In all triangle ABC holdsP a3
b+c ≥ 2√3sr.
Mihaly Bencze
PP28559. If a, b, c > 0 thenP 1
a +P a
b2+c2≥ (
�
a)2�
ab
P 1a+b ≥ 3
√3√
�
a2�
a
P 1a+b ≥ 3
P 1a+b .
Mihaly Bencze
PP28560. In all triangle ABC holds1).
P ab2+c2
≥ 10ss2+r2+2Rr
− s2+r2+4Rr4sRr
2).P ra
r2b+r2c
≥ (4R+r)2((4R+r)2+s2)4s4R
− 1r
3).P sin2 A
2
sin4 B2+sin4 C
2
≥ 8R(2R−r)2(16R2−24Rr+5r2+s2)(s2+r2−8Rr)((2R−r)(s2+r2−8Rr)−2Rr2)
− s2+r2−8Rrr2
4).P cos2 A
2
cos4 B2+cos4 C
2
≥ 4R(4R+r)2(5(4R+r)2+s2)(s2+(4R+r)2)((4R+r)3+s2(2R+r))
− (4R+r)2+s2
s2
Mihaly Bencze
PP28561. If a, b, c > 0 thenP a
b2+c2≥ 9abc
�
a−4(�
ab)2
4abc�
ab .
Mihaly Bencze
PP28562. In all triangle ABC holdsP a(brc+(c−b)ra−crb)
(bra+crb)(brc+cra)≥ (4R+r)2((4R+r)2+s2)
4s4R.
Mihaly Bencze
PP28563. In all triangle ABC holds 16s2r2 ≤P
a4 ≤ 4s2R2 (A newrefinement of Euler’s R ≥ 2r inequality).
Mihaly Bencze
PP28564. In all triangle ABC holdsP b+c−a
2(a−b)2+c2≥ s(s2+r2+4Rr)
8Rr2(4R+r)− 4R+r
2sr .
Mihaly Bencze
PP28565. In all triangle ABC holds�Pm2
am2bw
2c
�h2ah
2bh
2c ≤
�Pr2ar
2bh
2c
�w2aw
2bw
2c .
Mihaly Bencze
876 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP28566. In all triangle ABC holds�P
a2ma
� �Pam2
a
�≥ 72s3r3.
Mihaly Bencze
PP28567. In all triangle ABC holds
1).Q�
10ab+ 3p4 (a6 + b6)
�≤ 108s2
�s2 + r2 + 2Rr
�2
2).Q�
10rarb +3
q4�r6a + r6b
��≤ 432s4R2
Mihaly Bencze
PP28568. In all triangle ABC holds1).
P 1�
10ab+ 3√
4(a6+b6)≥ 5s2+r2+4Rr
2√3s(s2+r2+2Rr)
2).P 1
�
10rarb+3�
4(r6a+r6b)
≥ (4R+r)2+s2
4s2R√3
Mihaly Bencze
PP28569. In all triangle ABC holdsP tg4 A
2ctg4 C
2
18r2
s2ctgA
2ctgB
2−�
3rsctgC
2+2
≥ 3(s2−2r(4R+r))s2
.
Mihaly Bencze
PP28570. If ak > 0 (k = 1, 2, ..., n) , thenPr
10a1a2 +3
q4�a61 + a62
�≤ 2
√3
nPk=1
ak.
Mihaly Bencze
PP28571. If a, b > 0 then
49�a2
b2+ 2b
a
��b2
a2+ 2a
b
�≥ 225 + 90
�ab +
ba
�+ 9
�ab +
ba
�2.
Mihaly Bencze
PP28572. In all triangle ABC holds
� 4�
ctgA2
�
ctgA2ctgB
2
≥ 4
q3�rs
�11qPctgA
2 ctgB2 .
Mihaly Bencze
Proposed Problems 877
PP28573. In all triangle ABC holds 2s9r
�PqtgA
2
�2
≥P�
ctgA2+�
ctgB2
�
ctgC2
.
Mihaly Bencze
PP28574. If a, b ≥ 1 then(1+a)(1+b)(1+
√ab)(2+a+b)
(1+√a)(1+
√b)
≥ 2�√
a+√b�+ 2 (a+ b) .
Mihaly Bencze
PP28575. In all triangle ABC holds�P �tg3A2 ctg
2B2 ctg
5C2
� 14
��P �tg3C2 ctg
2B2 ctg
5A2
� 14
�≥ 27.
Mihaly Bencze
PP28576. In all triangle ABC holdsP 1
3�
ctg2 A2
�
3�
ctg2 B2− 3�
srtgA
2+ 3�
ctg2 C2
� ≥ 3rs
P3
qctgA
2 .
Mihaly Bencze
PP28577. If a, b > 0 then
1). a2+b2
2ab + ab(a+b)2
≥ 54
2). a2+b2
ab + 4ab(a+b)2
≥ 3
Mihaly Bencze
PP28578. If a, b, c > 0 then
9�P
a3� �P 1
a3
�+ 27 (
Pa)
�P 1a
�≥ 108 +
�P a+bc
�3.
Mihaly Bencze
PP28579. If a, b > 0 then 13 + 3a2+2ab+3b2
4(a2+ab+b2)≥ a(a+2b)
(a+b)(2a+b) +b(b+2a)
(a+b)(2b+a) .
Mihaly Bencze
PP28580. In all triangle ABC holdsPs
�
ctgA2
�
ctgB2+�
ctgC2
≥q
3r2s
PqctgA
2 .
Mihaly Bencze
878 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP28581. If a, b > 0 then 1+3(a6+b6)2a3b3
≥ a(a+2b)2a2+b2
+ 2a2+b2
a(a+2b) +b(b+2a)2b2+a2
+ 2b2+a2
b(b+2a) .
Mihaly Bencze
PP28582. If a, b, c > 0 thenP a
b+c ≥ 13
��
ab�
a2+
�
a2�
ab
�+ 5
6 ≥ 32 (A
refinement of Nesbit’s inequality).
Mihaly Bencze
PP28583. If a, b > 0 then a3+b3
a2+ab+b2+
2(a2+b2)√3(a2+ab+b2)
≥ a+ b.
Mihaly Bencze
PP28584. If a, b, c > 0 then
6
�2(a4+b4)a3+b3
+ a5+b5
2a2b2
�≥ (2a+b)4
(2a2+b2)(a+2b)+ (a+2b)4
(a2+2b2)(2a+b).
Mihaly Bencze
PP28585. In all triangle ABC holds
1).P r2
b
r3a3�
4(729r6+r6b)
≥ 118r3
2).P h2
b
h3a
3�
4(729r6+h6b)
≥ 118r3
Mihaly Bencze
PP28586. In all triangle ABC holds (4R+r)√3
s + k
q3k−
n2−1P �
ctgA2
�n ≥ 6
for all n, k ∈ N, n ≥ 2, k ≥ 6.
Mihaly Bencze
PP28587. If ak > 0 (k = 1, 2, ..., n) then
P a1a2+a3+...+an
+(n−1)n−1
n�
k=1ak
�
(a2+a3+...+an)≥ n+1
n−1 .
Mihaly Bencze
PP28588. If a, b > 0 then 4a2+ab+b2
2a + 4b2+ab+a2
2b + 4aba+b ≥ 4 (a+ b) .
Mihaly Bencze
Proposed Problems 879
PP28589. If ak > 0 (k = 1, 2, ..., n) then
P a1a2+a3+...+an
+
n�
k=1
ak�
(a2+a3+...+an)≥ n(n−1)n−1+1
(n−1)n.
Mihaly Bencze
PP28590. If a, b, c > 0 thenP a
b+c +4abc
�
(a+b) ≥ 2.
Mihaly Bencze
PP28591. If a, b, c > 0, then1).
P 1(a−b+c)2
+ 2P 1
ab ≥ 75(�
a)2
2).P 1
(a−b+c)2+ 1
2
P 1ab ≥ 27
(�
a)2
Mihaly Bencze
PP28592. If a, b, c > 0 then3P
a4 + 9P
a2b2 + 6abcP
a ≥ 6 (P
ab)�P
a2�+ (P |(a− b) (b− c)|)2 .
Mihaly Bencze
PP28593. If ak > 0 (k = 1, 2, ..., n) thenP a21a
22(a21+a22)
a1a2+�
2(a41+a42)+ 1
3
nPk=1
a4k ≥ 2P
a21a22.
Mihaly Bencze
PP28594. If a, b > 0 then a�
2√a2+ab+b2
+ 1a√3
�2+ b
�2√
a2+ab+b2+ 1
b√3
�2≥
≥ 1
(a+ b)2 (a+ 2b) (2a+ b)
�5a2 + 11ab+ 2b2
�2
2a+ b+
�5b2 + 11ab+ 2a2
�2
a+ 2b
!.
Mihaly Bencze
PP28595. If a, b, c > 0 thenP
a3 ≥ 3abc+P
(a− b)2 c.
Mihaly Bencze
PP28596. If 1 < a, b ≤ 103 then
8�
a(a−1)2
+ b(b−1)2
�≥ (10− 2a− b)2 + (10− a− 2b)2 .
Mihaly Bencze
880 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP28597. If a, b, c > 0 then 1 +�
a2
2�
ab ≤P ab+c ≤ 1
2 +�
a2�
ab (A new
refinement of the inequalityP
a2 ≥P ab).
Mihaly Bencze
PP28598. If a, b, c > 0 and λ > 0 thenP a(b+c)λ+1
(b2+bc+c2)λ≥ 2λ+1
�
ab
(�
a)λ.
Mihaly Bencze
PP28599. If a, b > 0 then�a4 + a3b+ b4
� �a4 + ab3 + b4
�≥
≥�2a2 + b2 + 6 (a− b)2
��a2 + 2b2 + 6 (a− b)2
�a2b2.
Mihaly Bencze
PP28600. If a, b > 0 then(4a2+ab+b2)(a2+ab+4b2)
4ab(a+b)2≤
�12 + 2a2+b2
a(a+2b)
��12 + a2+2b2
b(b+2a)
�.
Mihaly Bencze
PP28601. If a, b > 0 then4a2b2(1+a2)(1+b2)
(1+a)(1+b) +a4(1+a2)
2
(1+a)2+
b4(1+b2)2
(1+b)2≥ 8
�3− 2
√2�ab
�a2 + ab+ b2
�.
Mihaly Bencze
PP28602. If ak > 0 (k = 1, 2, ..., n) , thenP
1≤i<j≤n
a4i+a4jai+aj
≥ (n− 1)nP
k=1
a2k + 3P
1≤i<j≤n(ai − aj)
2 .
Mihaly Bencze
PP28603. If a, b > 0 and λ > 0 then�a2λ+2 + b2λ+2 + aλ+2bλ
� �a2λ+2 + b2λ+2 + aλbλ+2
�≥ 9(2a3+b3)(a3+2b3)a2λb2λ
(2a+b)(a+2b) .
Mihaly Bencze
PP28604. If a, b > 0 then�2p
(a2 + 2) (2a+ 1) +p(b2 + 2) (a+ b+ 1)
�2+
+�2p(b2 + 2) (2b+ 1) +
p(a2 + 2) (a+ b+ 1)
�2≥ 18 (a+ b+ 1)2 .
Mihaly Bencze
Proposed Problems 881
PP28605. In all tetrahedron ABCD holds1).
P 3ha+2λrha+λr ≥ 8(λ+6)
λ+4
2).P 3ra+2λr
ra+λr ≥ 8(λ+3)λ+2 for all λ > 0
Mihaly Bencze
PP28606. If x, y, z > 0 thenP x
y+z +Pq
x+y+2z2(x+y) ≥ 9
2 .
Mihaly Bencze
PP28607. If ak > 1 (k = 1, 2, ..., n) and 0 < x ≤ y thennP
k=1
�ayk − axk
�≥ (y − x)
�nP
k=1
ax+y2
k ln ak
�.
Mihaly Bencze
PP28608. Let be P (x) = a0 + a1x+ ...+ amxm, λi > 0 (i = 1, 2, ..., n) ,nQ
i=1λi = 1 then
nPi=1
P k (λi) ≥ nP k (1) , for all k ∈ N .
Mihaly Bencze
PP28609. If ak > 0 (k = 1, 2, ..., n) then
2n(n−1)
�1n
nPk=1
ak
�n2
≥�
nQk=1
ak
� Q1≤i<j≤n
(ai + aj)2
!.
Mihaly Bencze
PP28610. If a, b, c > 1 and 0 < x ≤ y then
ay + by + cy − ax − bx − cx ≥ (y − x)�a
x+y2 ln a+ b
x+y2 ln b+ c
x+y2 ln c
�.
Mihaly Bencze
PP28611. In all triangle ABC holds 4P
ctg2A2 + 45s4
81r4+ 2r(4R+r)
s2≥ 19.
Mihaly Bencze
PP28612. If a, b, c, d > 0 then(P
a)16 ≥ 268
516abcd (a+ b)2 (a+ c)2 (a+ d)2 (b+ c)2 (b+ d)2 (c+ d)2 .
Mihaly Bencze
882 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP28613. If xk > 0 (k = 1, 2, ..., n) andnP
k=1
xk = n thenP x3
1
x2
√x31+8
≥ n3 .
Mihaly Bencze
PP28614. If xk > 0 (k = 1, 2, ..., n) andnP
k=1
xk = n then
P xλ+11
x2
�
xλ+12 +λ2+2λ
≥ nλ+1 .
Mihaly Bencze
PP28615. If a, b > 0 then6(a2+ab+b2)
a+b +3(a3+b3)
2ab ≥ 514 (a+ b) +
32(a−b)2(a2+ab+b2)ab(a+b)2(2a+b)(a+2b)
.
Mihaly Bencze
PP28616. In all triangle ABC holds
1). 4P 1
r4a+ 45
81r4≥ 19((4R+r)2−2s2)
s4r2
2). 4P 1
h4a+ 45
81r4≥ 19
��s2+r2+4Rr
4s2r2
�2− R
s2r3
�
Mihaly Bencze
PP28617. If 0 < a < b < c then (c− a) (b− a) arctg c−b1+cb ≤
(c− b) (b− a) arctg c−a1+ca ≤ (c− b) (c− a) arctg b−a
1+ab .
Mihaly Bencze
PP28618. If a, b, c > 0 then 3P 2a2+bc
b+c ≥ 92
Pa+ 16abc
�
(a−b)2
(�
a)�
(a+b) .
Mihaly Bencze
PP28619. If 0 < a1 ≤ a1 ≤ a2 ≤ ... ≤ an thenQ
1≤i<j≤n
eaj−aaiaj−ai
≥ en−12
n�
k=1ak.
Mihaly Bencze
PP28620. If e ≤ a ≤ b ≤ c then
ln�cb
�c−aln (cb)b−a ≤ ln
�ca
�c−bln (ac)b−a ≤ ln
�ba
�c−bln (ab)c−a . If
0 < a ≤ b ≤ c ≤ e then holds the reverse inequalities.
Mihaly Bencze
Proposed Problems 883
PP28621. If 0 < a ≤ b ≤ c then(b− a) (c− a) (c− b) ea+b+c ≤
�eb − ea
�(ec − ea)
�ec − eb
�.
Mihaly Bencze
PP28622. If 0 < a ≤ b < 1 thenb�ln2 b− 2 ln b+ b
�− a
�ln2 a− 2 ln a+ a
�≥
≥ b2 − a2
2
�ln2
�a+ b
2
�− 2 ln
�a+ b
2
�+
a+ b
2
�.
Mihaly Bencze
PP28623. If 1 < a < 1 + 1n then
nQk=1
�ak
k − ak+1
k+1
�< 1
(n+1)(n!)2.
Mihaly Bencze
PP28624. If a > 1 thennQ
k=1
�ak − 1
k+1
�< a
n(n+3)2
(n+1)(n!)2.
Mihaly Bencze
PP28625. Denote Lk the kth Lucas number. Prove that
1).�1 + 1
m
�(Ln+2 − 3) <
nPk=1
Lm+1k+1 −Lm+1
k
Lmk+1−Lm
k<
�1 + 1
m
�(Ln+3 − 4)
2).�1 + 2
m
�(LnLn+1 − 2) <
nPk=1
Lm+2k+1 −Lm+2
k
Lmk+1−Lm
k<
�1 + 2
m
�(Ln+1Ln+2 − 3) for
all m ∈ N∗
Mihaly Bencze
PP28626. Denote Fk the kth Fibonacci number. Prove that
1).�1 + 1
m
�(Fn+2 − 1) <
nPk=1
Fm+1k+1 −Fm+1
k
Fmk+1−Fm
k<
�1 + 1
m
�(Fn+3 − 2)
2).�1 + 2
m
�FnFn+1 <
nPk=1
Fm+2k+1 −Fm+2
k
Fmk+1−Fm
k<
�1 + 2
m
�(Fn+1Fn+2 − 1) for all
m ∈ N∗
Mihaly Bencze
PP28627. In all triangle ABC holds 6qtgA
2 −qtgB
2 +qtgC
2 < 3p
sr .
Mihaly Bencze
884 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP28628. If ai > 0 (i = 1, 2, ..., n) thennP
i=1ai ≥ k+n−1
snQ
i=1ai
�nP
i=1
k+n−1
qak−1i
�when k ∈ N∗.
Mihaly Bencze
PP28629. If ak ≥ 2 (k = 1, 2, ..., n) and A = 1n
nPk=1
ak, G = n
snQ
k=1
ak,
H = nn�
k=1
1ak
, then A3G3H5 +A3G3 ≤ A3G5H3 +A3H3 ≤ A5G3H3 +G3H3.
Mihaly Bencze
PP28630. In all triangle ABC holds1). 6
√ra −
√rb +
√rc <
3s√r
2). 6√ha −
√hb +
√hc < 3s
q2R
Mihaly Bencze
PP28631. Compute limn→∞
n
�1−
n+1Rn
e1xdx
�.
Mihaly Bencze
PP28632. If a, b, c > 0 then
3P �
a2 − bc�4
+ 3P
a2 (b+ c)4 ≥��P
a2�2
+P
a2b2�.
Mihaly Bencze
PP28633. Solve in R+ the following system:
4πarctgx1 +�arcsin 2x2
1+x22
�2= π2 + 4arctg2x3
4πarctgx2 +�arcsin 2x3
1+x23
�2= π2 + 4arctg2x4
−−−−−−−−−−−−−−−−−−−−4πarctgxn +
�arcsin 2x1
1+x21
�2= π2 + 4arctg2x2
.
Mihaly Bencze
PP28634. Solve in Z the equation x2+(y+z)2
(x+y)2+z2+ y2+(z+x)2
(y+z)2+x2+ z2+(x+y)2
(z+x)2+y2= 3.
Mihaly Bencze
Proposed Problems 885
PP28635. Solve in R the following system: arcsin 2x1
1+x21+ 2arctgx2 =
arcsin 2x2
1+x22+ 2arctgx3 = ... = arcsin 2xn
1+x2n+ 2arctgx1 = π.
Mihaly Bencze
PP28636. Prove that
limx→0
1x
�3
nPk=1
qk2 (k + 1)2 + x− n (n+ 1) (n+ 2)
�= 3n
2(n+1) .
Mihaly Bencze
PP28637. Prove that1R0
x2 arcsin 2xx2+1
dx = π6 + 1
2 − ln 2. Compute
In =1R0
xn arcsin 2xx2+1
dx.
Mihaly Bencze
PP28638. Solve in Z the equation xy2z3
x+2y+3z + yz2x3
y+2z+3x + zx2y3
z+2x+3y = 12 .
Mihaly Bencze
PP28639. If a, b, c, p, q > 0 thenpa+qbp+qR0
ex ln�x2 + x+ c
�dx+
pb+qap+qR0
ex ln�x2 + x+ c
�dx ≤
≤aZ
0
ex ln�x2 + x+ c
�dx+
bZ
0
ex ln�x2 + x+ c
�dx.
Mihaly Bencze
PP28640. In all triangle ABC holds 1 ≤� 1
a2� 1
ab
≤ R2r (A new refinement of
Euler’s R ≥ 2r inequality).
Mihaly Bencze
PP28641. If x ∈�0, π2
�then compute I1 =
Rtg (arccos (sin (arctgx))) dx,
I2 =Rtg (arcsin (cos(arcctgx))) dx.
Mihaly Bencze
886 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP28642. Prove that limn→∞
1n
nPk=0
n√akbn−k = a−b
ln a−ln b when a > b > 0.
Mihaly Bencze
PP28643. The triangle ABC is equilateral if and only if
2 (cosA)3 |cosB|+ 2 (cosB)3 |cosA| = (1 + cos 2C)�cos2A+ cos2B
�
2 (cosB)3 |cosC|+ 2 (cosC)3 |cosB| = (1 + cos 2A)�cos2B + cos2C
�
2 (cosC)3 |cosA|+ 2 (cosA)3 |cosC| = (1 + cos 2B)�cos2C + cos2A
� .
Mihaly Bencze
PP28644. If x, y, z ∈ R∗+, show that
x2 + 1
2y + 3z+
y2 + 1
2z + 3x+
z2 + 1
2x+ 3y≥ 6
5and
find the values for which the equality is achieved.
Mihaela Berindeanu
PP28645. For a nonnegative integer n let Fn be the nth Fibonacci numberdefined by F0 = 0, F1 = 1 and for all n ≥ 2, Fn = Fn−1 + Fn−2. Show that
1
54F2n
������
4Fn+1 Fn Fn + 2Fn−1
FnFn+1 (2Fn+1 + Fn−1)2 F2n
F 2n+1 + Fn−1Fn+1 F2n F 2
n+2
������
Jose Luis Dıaz-Barrero
PP28646. Let a, b be positive integers. A frog staying at point (a, b) of theplane may jump to point (a− b, b) if a > b or to point (a, b− a) if a < b. Forinstance a valid path for a frog starting at point (17, 12) is the following:(17, 12) → (5, 12) → (5, 7) → (5, 2) → (3, 2) → (1, 2) → (1, 1) Currently, thefrog is at point P (20172017, 123456). It is possible after some jumps to arriveat point (1, 1)?
Jose Luis Dıaz-Barrero
PP28647. Let a, b, c be three positive numbers such that
ab+ bc+ ca = 3abc. Prove thatq
a+bc(a2+b2)
+q
b+ca(b2+c2)
+q
c+ab(c2+a2)
≤ 3.
Jose Luis Dıaz-Barrero
Proposed Problems 887
PP28648. In all acute triangle ABC holdsQ
(cosA+ cosB cosC)cosA ≥�R+r2R
�R+rR .
Mihaly Bencze
PP28649. In all acute triangle ABC holdsQ
(sinB sinC)cosA ≥�R+r2R
�R+rR .
Mihaly Bencze
PP28650. Prove that 2nP
k=1
(√k2+2k+2−
√k2+1)
2
2k+1 < ln n+1+√n2+2n+2
1+√2
.
Mihaly Bencze
PP28651. ProvenP
k=1
�2k+32k+5
�2k+5(k+3)k+3
(k+1)k+2 ≥ n(n+5)2 .
Mihaly Bencze
PP28652. Denote Lk the kth Lucas number, prove that
nPk=1
�4√Lk +
4pLk+1 +
4pLk+3
�4pLkLk+1Lk+3 ≤ Ln+6 − 13.
Mihaly Bencze
PP28653. In all triangle ABC holds√2sr ≤ min {ab, bc, ca} .
Mihaly Bencze
PP28654. In all triangle ABC holds1).
P a(ha−2r)ha+2r = 2sr(12R+r)
4s2+r2+8Rr
2).P (b+c−a)(ha−2r)
ha+2r =2s(4s2−3r2−40Rr)
4s2+r2+8Rr
Mihaly Bencze
PP28655. Denote Fk the kth Fibonacci number, prove thatnP
k=1
�4√Fk +
4pFk+1 +
4pFk+3
�4pFkFk+1LF ≤ Fn+6 − 8.
Mihaly Bencze
888 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP28656. If x ≥ 0 then
�1 +
nPk=1
xk
k!
�n+1
≤�1 +
n+1Pk=1
xk
k!
�n
.
Mihaly Bencze
PP28657. If a ∈ (−1, 1) then limn→∞
nQk=0
�1 + a4
k+ a2·4
k+ a3·4
k�= 1
1−a .
Mihaly Bencze
PP28658. Prove that:
1). limn→∞
�1 +
nPk=1
1k!
� 1n
= 1
2). limn→∞
n
�1 +
nPk=1
1k!
� 1n
− 1
!= 1
Mihaly Bencze
PP28659. Prove that
�1 +
nPk=1
1k!
�n+1
<
�1 +
n+1Pk=1
1k!
�n
for all n ∈ N∗.
Mihaly Bencze
PP28660. If a1 = 1 and ak =
r2
q3p4...
√k for k ≥ 2 then compute
limn→∞
1n2
√n
nPk=1
a2k.
Mihaly Bencze
PP28661. Prove that limn→∞
nPk=1
arctg 2525k2+55k+49
= π2 − arctg 8
5 .
Mihaly Bencze
PP28662. Prove that 1!2!...n! < 22n+1−2 for all n ≥ 1.
Mihaly Bencze
PP28663. If a1 = 1 and ak =
s
2
r3
q4p
5...√k for k ≥ 2 then
1).nP
k=1
�akak+1
k(k+1)
�2< 4n
n+1
2).nP
k=1
a2k ≤ n (n+ 1)
Proposed Problems 889
3).nP
k=1
1a2ka2k+1
> n4(n+1)
Mihaly Bencze
PP28664. If n ≥ 2, n ∈ N then solve in R the following system:
nnx1 + n− 1 = nx2+1
nnx2 + n− 1 = nx3+1
−−−−−−−−−nnxn + n− 1 = nx1+1
.
Mihaly Bencze
PP28665. Prove thatnP
k=1
(k!)−2k < 2n.
Mihaly Bencze
PP28666. If a, b > 1 then 8a�ln2 a− ln a2
�+ 8b
�ln2 b− ln b2
�≥
(3a+ 5b)�ln2
�3a+5b
8
�− ln
�3a+5b
8
�2�+ (5a+ 3b)
�ln2
�5a+3b
8
�− ln
�5a+3b
8
�2�.
Mihaly Bencze
PP28667. Solve in R the following system:
27x+y + 2 = 3y+z+1
27y+z + 2 = 3z+x+1
27z+x + 2 = 3x+y+1.
Mihaly Bencze
PP28668. If x ∈�0, π2
�then compute
Rsinx cosx−x−x2 sinx
x(x+sinx) cosx dx.
Mihaly Bencze
PP28669. If 1 < a ≤ b then
2π ln b
a ln ab+ 2 ln�be
�b � ea
�a<
bRa
lnxdxarctgx < π
2 lnba ln ab+ 2 ln
�be
�b � ea
�a.
Mihaly Bencze
PP28670. In all triangle ABC holds
27r3
s3
Pctg4A2 +
�PqctgA
2
�2
≥ 36(4R+r)(s2−2r(4R+r))s3
.
Mihaly Bencze
890 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP28671. In all triangle ABC holds
1).P�
rbra
�2≥ 81r2((4R+r)2−2s2)
s4
2).P�
hb
ha
�2≥ 81r4
��s2+r2+4Rr
4s2r2
�2− R
s2r3
�
Mihaly Bencze
PP28672. In all triangle ABC holds
1). 27r3P 1
r4a+�P 1√
ra
�2≥ 36(4R+r)(s2−2r(4R+r))
s4
2). 27r3P 1
h4a+�P 1√
ha
�2≥ 9(s2−r2−4Rr)(s2+r2+4Rr)
2s2r
Mihaly Bencze
PP28673. In all triangle ABC holds
1).Pq
1 + rbra
≥q
23 · s
r
2).Pq
1 + hb
ha≥ 2s
r
q2r3R
Mihaly Bencze
PP28674. In all triangle ABC holdsP
ctgA2
qctgA
2 + ctgB2 ≥
�sr
� 32 .
Mihaly Bencze
PP28675. If a, b > 0 then a2+2b2
b(2a+b) +b2+2a2
a(2b+a) +8ab
(a+b)2≥ 4.
Mihaly Bencze
PP28676. If a, b > 0 then2
a2(2a+b)2+ 2
b2(2b+a)2+ 1
(a2+2b2)2+ 1
(b2+2a2)2≥ 1
9
�(2a+b)2
2a6+b6+ (a+2b)2
a6+2b6
�.
Mihaly Bencze
PP28677. In all triangle ABC holdsrs2
P (rctgA2+stgA
2 )ctg2 A
2
tgB2+tgC
2
+P ctgA
2
tgB2+2tgC
2
≥ 4 + rs
P ctg2 A2 (tg
B2−tgA
2 )(tgC2−tgA
2 )tgB
2+tgC
2
.
Mihaly Bencze
PP28678. If a, b, c > 0 and λ > 0 thenP a
(b2+bc+c2)λ≥
�
a
(�
ab)λ.
Mihaly Bencze
Proposed Problems 891
PP28679. In all triangle ABC holds
1).P�
r(r2a+rbrc)r2a(rb+rc)
+ rbrcr(ra+2rb)
�≥ 4
P (rb−ra)(rc−ra)r2a
2).P�
r(h2a+hbhc)
h2a(hb+hc)
+ hbhc
r(ha+2hb)
�≥ 4
P (hb−ha)(hc−ha)h2a
Mihaly Bencze
PP28680. If ak > 0 (k = 1, 2, ..., n) , thenP �
a21 − a1a2 + a22�r
10a1a2 +3
q4�a61 + a62
�≤ 2
√3
nPk=1
a3k.
Mihaly Bencze
PP28681. If a, b, c > 0 thenP a
b2+bc+c2≥
�
a�
ab .
Mihaly Bencze
PP28682. Solve in R the equation [xn!] + [x (n+ 1)!] = (n+ 2)! where [·]denote the integer part, and n ∈ N.
Mihaly Bencze
PP28683. Solve in R the equationh
xn+2
i+n
xn+1
o= 1
n where [·] and {·}denote the integer , respective the fractional part and n ∈ N∗.
Mihaly Bencze
PP28684. Solve in R the equation [2017x] + [2018x] = 2019 where [·] denotethe integer part.
Mihaly Bencze
PP28685. The triangle ABC is equilateral if and only if32 +
P�ma
wa
�2+P a+b
c =P (b+c)ma√
bcwa.
Mihaly Bencze
PP28686. Determine all a, b ∈ N and all prime p for which(p+ 2018)a = (p− 2018)b .
Mihaly Bencze
892 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP28687. Determine all prime p, q, r for which
p3q3r3 + 15p3 + 5q3 + 3r3 = 5p3q3 + q3r3 + 3r3p3 + 120.
Mihaly Bencze
PP28688. In all triangle ABC holdsP
ctg6A2 + 6�sr
�2 ≤ sr
Pctg5A2 .
Mihaly Bencze
PP28689. Let be A =
�a ba−a2−1
b 1− a
�where a, b ∈ C, b 6= 0. Prove that
1). A2n is invertable and�A2n
�−1= A2n−1
for all n ∈ N,n ≥ 2
2). det
�6n+6Pk=0
(−1)k Ak+1
�=
6n+6Pk=0
(−1)k det�Ak+1
�
Mihaly Bencze
PP28690. Solve in Z the equation y8 − x (x+ 1) (x+ 2) (x+ 3) = 136.
Mihaly Bencze
PP28691. In all triangle ABC holds1). r
P 1r6a
+ 6s2r
≤P 1r5a
2). rP 1
h6a+ 3R2
2s4r≤P 1
h5a
Mihaly Bencze
PP28692. If ak > 1 (k = 1, 2, ..., n) , thenP
loga1aλ+12 +aλ+1
3
a22+a23≥ n (λ− 1) for
all λ ≥ 1.
Mihaly Bencze
PP28693. Let ABC be a triangle, denote AD,BE,CF the interiorbisectors, where D ∈ (BC) , E ∈ (CA) , F ∈ (AB) . If one of Area [ABD] ,Area [BCE] , Area [CAF ] is the arithmetical mean of the other two, thenthe triangle ABC is isoscel.
Mihaly Bencze
Proposed Problems 893
PP28694. If a, b, c > 0 and a+ b+ c = 3 then(n+1)(
�
a2)2
�
a2b+ 2n
�
ab�
a2≥ n (3n+ 5) for all n ∈ N∗.
Mihaly Bencze
PP28695. If a, b, c > 0 thenP a2c(b+c)√
(a2+b2)(a2+c2)≤P ab.
Mihaly Bencze
PP28696. Solve in Z the equation 1x + 1
x+y + 1x+y+z = 1.
Mihaly Bencze
PP28697. Let (an)n≥1 be an arithmetical progression. Prove that:
1). 1(a1+a2)(a2+a3)
+ 1(a2+a3)(a3+a4)
+ ...+ 1(an+an+1)(an+1+an+2)
=
= n(a1+a2)(an+1+an+2)
2). 1(a1+a2)(a2+a3)(a3+a4)
+ 1(a2+a3)(a3+a4)(a4+a5)
+ ...
+ 1(an+an+1)(an+1+an+2)(an+2+an+3)
= n(a1+a2)(a2+a3)(an+1+an+2)(an+2+an+3)
Mihaly Bencze
PP28698. In all acute triangle ABC holdsP 1
2+tg2A≥ 2(R+r)2
12R2+4Rr+r2−s2.
Mihaly Bencze
PP28699. In all triangle ABC holdsP 1
2+ctg2A≥ 2s2
6R2−4Rr−r2+s2.
Mihaly Bencze
PP28700. In all triangle ABC holds
1).Q �
r2a + 9r2�≤ 8s6
27
2).Q �
h2a + 9r2�≤ 8s6
27
Mihaly Bencze
PP28701. In all triangle ABC holdsQ �
1 + 9tg2A2 tg2B2
�≤ 8s2
27r2.
Mihaly Bencze
894 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP28702. In all triangle ABC holds
1).�4R+r4R
�2 ≥ 3�
s4R
�2+ 3
16
�P ��sin2 A2 − sin2 B
2
���2
2).�s2+r2−8Rr
16R2
�2≥ 3r2(2R−r)
32R3 + 316
�P ��sin2 A2 − sin2 C
2
�� sin2 B2
�2
3).�4R+r4R
�2 ≥ 3�
s4R
�2+ 3
16
�P ��cos2 A2 − cos2 B
2
���2
4).�s2+(4R+r)2
16R2
�2≥ 3s2(4R+r)
32R3 + 316
�P ��cos2 A2 − cos2 C
2
�� cos2 B2
�2
Mihaly Bencze
PP28703. In all triangle ABC holds1). s2 ≥ 3r (4R+ r) + 3
16 (P |a− b|)2
2).�s2 + r2 + 4Rr
�2 ≥ 20s2Rr + 316 (P |a− c| b)2
3).�s2+r2+4Rr
2R
�2≥ 6s2r
R + 316 (P |ha − hb|)2
4). (4R+ r)2 ≥ 3s2 + 316 (P |ra − rb|)2
5). s4 ≥ 3s2r (4R+ r) + 316 (P |ra − rc| rb)2 (New refinements of some
classical inequalities).
Mihaly Bencze
PP28704. If λ ∈ (0, 1] thennP
k=1
k12λ ≤ n1−λ (n+ 1)
3λ2 .
Mihaly Bencze
PP28705. If ak > 0 (k = 1, 2, ..., n) and λ ∈ [0, 1] thennP
k=1
�ak+1
a2k+2ak+3
�λ≤ n · 2−3λ
2 .
Mihaly Bencze
PP28706. If 0 < a < b and ab = 4 then computeln bRln a
xe3xdx(ex+2)6
.
Mihaly Bencze
PP28707. Prove thatnP
k=1
1k ≥ 4n
(√n+1)
2 .
Mihaly Bencze
Proposed Problems 895
PP28708. Prove thatnP
k=1
(n−k)(n−k+2)
(V kn )
2 = 1− 1(n!)2
.
Mihaly Bencze
PP28709. Prove thatnP
k=1
(n−k)((n−k)2+3(n−k)+3)(V k
n )3 = 1− 1
(n!)3.
Mihaly Bencze
PP28710. ComputeR (x+1)2(x+5)dx
x6+6x5+15x4+20x3+24x2+19x+2where x > 0.
Mihaly Bencze
PP28711. In all tetrahedron ABCD holds
1).P�
rha
�har
< 7
2).P�
r2ra
� 2rar
< 7
Mihaly Bencze
PP28712. Prove thatnP
k=1
n−kV kn
= 1− 1n! .
Mihaly Bencze
PP28713. Prove thatnP
k=1
�n+1
kn(k+1)
� kn(k+1)n+1
< 2n+ 1.
Mihaly Bencze
PP28714. In all triangle ABC holds
1).P�
rra
� rar< 5
2).P�
rha
�har
< 5
Mihaly Bencze
PP28715. Determine all triangles ABC in which (x+ y) a = xb+ yc for allx, y > 0.
Mihaly Bencze
896 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP28716. If a, b, c, d ∈ N ∗ then�a+ b1+
1b
��c+ d1+
1d
�≤
�a+ ba
1a
��c+ dc
1c
�.
Mihaly Bencze
PP28717. Let be A ∈ M2 (R) where A =
�a bc d
�. Determine all λ ∈ R
and m > 0 such that aλ + bλ + cλ + dλ < m and I2 +A is invertible.
Mihaly Bencze
PP28718. In all triangle ABC holdsP a+b
c5≥
��s2+r2+4Rr
4sRr
�2− 1
Rr
�2
.
Mihaly Bencze
PP28719. In all triangle ABC holds
1).P ra+rb
r5c≥
�s2−2r(4R+r)
s2r2
�2
2).P rarb+rbrc
r5cr5a
≥�(4R+r)2−2s2
s4r2
�2
3).P a
(s−a)5≥
�(4r+r)2−2s2
s2r2
�2
4).P c(s−b)
(s−a)5(s−c)5≥
�s2−2r2−8Rr
s2r4
�2
5).P ha+hb
h5c
≥�s2−r2−4Rr
2s2r2
�2
6).P hahb+hbhc
h5ch
5a
≥��
s2+r2+4Rr4s2r2
�2− R
s2r3
�2
Mihaly Bencze
PP28720. Let ABC be a triangle in which A ≤ B ≤ C ≤ π2 . Prove that�
s2+r2+Rr2sr
�2− 4R
r +P A cosA
sin2 A≤ 9(A+C)2
4πAC .
Mihaly Bencze
PP28721. If ai > 0 (i = 1, 2, ..., n) and k ∈ {2, 3, ..., n} , thenQ
cyclic
�a1ak
+ a2ak
+ ...+ak−1
ak+ 1
�ak ≤ k
n�
i=1ai.
Mihaly Bencze
Proposed Problems 897
PP28722. If x, y, z ≥ 1 then
P 1x + 2 lnxyz + 9
x+y+z + 6 ln x+y+z3 ≤ 4
P 1x+y + 4
Pln x+y
2 .
If x, y, z ∈ (0, 1) then holds the reverse inequality.
Mihaly Bencze
PP28723. In all acute triangle ABC holdsR+rR + 2 ln 4R2
s2−(2R+r)2+
9(s2−(2R+r)2)s2+r2−4R2 + 6 ln s2−(2R+r)2
3(s2+r2−4R2)≥
≥�s2 + r2 − 4R2
�2+ 4R (R+ r)
�s2 − (2R+ r)2
�
Rr (s2 + r2 + 2Rr)+4 ln
�s2 − (2R+ r)2
�2
32Rr (s2 + r2 + 2Rr).
Mihaly Bencze
PP28724. In all acute triangle ABC holdss2+r2−4R2
s2−(2R+r)2+ 2 ln s2−(2R+r)2
4R2 + 9RR+r + 6 ln R+r
3R ≤
≤ 4R�s2 + 5r2 + 8Rr
�
r (s2 + r2 + 2Rr)+ 4 ln
r�s2 + r2 + 2Rr
�
32R3.
Mihaly Bencze
PP28725. In all triangle ABC holdss2+(4R+r)2
s2+ 4 ln s
4R + 18R4R+r + 6 ln 4R+r
6R ≤
≤8R
�5 (4R+ r)2 + s2
�
(4R+ r)3 + s2 (2R+ r)+ 4 ln
(4R+ r)3 + s2 (2R+ r)
256R3.
Mihaly Bencze
PP28726. In all triangle ABC holds4R+r2R + 4 ln 4R
s + 9s2
s2+(4R+r)2+ 6 ln s2+(4R+r)2
3s2≥
≥ 4X cos2 A
2 cos2 B2
cos2 A2 + cos2 B
2
+ 4 lns4
64R�(4R+ r)3 + s2 (2R+ r)
� .
Mihaly Bencze
898 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP28727. In all triangle ABC holdss2+r2−8Rr
r2+ 4 ln r
4R + 18R2R−r + 6 ln 2R−r
6R ≤
≤ 8R�16R2 − 24Rr + 5r2 + s2
�
(2R− r) (s2 + r2 − 8Rr)− 2Rr2+ 4 ln
(2R− r)�s2 + r2 − 8Rr
�− 2Rr2
256R3.
Mihaly Bencze
PP28728. In all triangle ABC holds2R−r2R + 4 ln 4R
r + 9r2
s2+r2−8Rr+ 6 ln s2+r2−8Rr
3r2≥
≥ 4X sin2 A
2 sin2 B2
sin2 A2 + sin2 B
2
+ 4 lnr4
64 ((2R− r) (s2 + r − 8Rr)− 2Rr2).
Mihaly Bencze
PP28729. In all triangle ABC holdssR + 2 ln 2R2
sr + 18srs2+r2+4Rr
+ 6 ln s2+r2+4Rr6sr ≥
≥�s2 + r2 + 4Rr
�2 − 8sRr
sR (s2 + r2 + 2Rr)+ 4 ln
r2
8R (s2 + r2 + 2Rr).
Mihaly Bencze
PP28730. In all triangle ABC holdss2+r2+4Rr
2sr + 2 ln sr2R2 + 9R
s + 6 ln s3R ≤ 4R(5s2+r2+4Rr)
s(s2+r2+2Rr)+ 4 ln
s(s2+r2+2Rr)32R3 .
Mihaly Bencze
PP28731. Prove that1R0
[nx] cos (a [nx]) dx = n cos(n−1)a−(n−1) cosna−1
4n sin2 a2
, where
n ∈ N∗.
Mihaly Bencze and Ovidiu Furdui
PP28732. Prove that1R0
arctg 1([nx])2+3[nx]+3
dx = 1narctg
n−1n+1 where [·] denote
the integer part, and n ∈ N∗.
Mihaly Bencze and Ovidiu Furdui
Proposed Problems 899
PP28733. Prove that 12 +
1R0
cos (2a [nx]) dx = sin(2n−1)a2n sin a , where [·] denote
the integer part, and n ∈ N∗.
Mihaly Bencze and Ovidiu Furdui
PP28734. Prove that1R0
[nx] sin (a [nx]) dx = n sin(n−1)a−(n−1) sinna
4n sin2 a2
, where [·]denote the integer part, and n ∈ N∗.
Mihaly Bencze and Ovidiu Furdui
PP28735. If a > 1 then1R0
a[nx]dx = an−1n(a−1) , when [·] denote the integer part,
and n ∈ N∗.
Mihaly Bencze and Ovidiu Furdui
PP28736. Prove that1R0
sin (2a [nx]) dx = sinna sin(n−1)an sin a , when [·] denote the
integer part, and n ∈ N∗.
Mihaly Bencze and Ovidiu Furdui
PP28737. Prove that1R0
[nx]dx
4([nx])4+1= n−1
2(2n2−2n+1), where n ∈ N∗.
Mihaly Bencze and Ovidiu Furdui
PP28738. Prove that1R0
ln�[nx]+2[nx]+1
�dx = 1
n ln (n+ 1) , when [·] denote the
integer part, and n ∈ N∗.
Mihaly Bencze and Ovidiu Furdui
PP28739. Prove that1R0
[nx] ([nx])!dx = n!−1n , when [·] denote the integer
part, and n ∈ N∗.
Mihaly Bencze and Ovidiu Furdui
900 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP28740. Prove that1R0
([nx])!�([nx])2 + [nx] + 1
�dx = n·n!−1
n , when [·]denote the integer part, and n ∈ N∗.
Mihaly Bencze and Ovidiu Furdui
PP28741. Prove that1R0
dx([nx]+1)([nx]+2) =
1n+1 , when [·] denote the integer
part, and n ∈ N∗.
Mihaly Bencze and Ovidiu Furdui
PP28742. Prove that1R0
[nx]dx([nx]+1)! =
�1− 1
n!
�1n , when [·] denote the integer
part, and n ∈ N∗.
Mihaly Bencze and Ovidiu Furdui
PP28743. If 0 ≤ x1 ≤ x2 ≤ ... ≤ xn ≤ 1 then(x2 − x1) arctgx1 + (x3 − x1) arctgx2 + (x4 − x2) arctgx3 + ...+(xn − xn−2) arctgxn−1 + (xn − xn−1) arctgxn ≤ π
2 − ln 2.
Mihaly Bencze and Sitaru Daniel
PP28744. In all tetrahedron ABCD holds1). 1
r2a+ 1
r2b
+ 1r2c
+ 1r2d
+ 8r2
rarbrcrd≥ 1
rarb+ 1
rarc+ 1
rard+ 1
rbrc+ 1
rbrd+ 1
rcrd
2). 1h2a+ 1
h2b
+ 1h2c+ 1
h2d
+ 32r2
hahbhchd≥ 1
hahb+ 1
hahc+ 1
hahd+ 1
hbhc+ 1
hbhd+ 1
hchd
Mihaly Bencze
PP28745. In all acute triangle ABC holdsP
A (π − 2A) ≤ srπ2
2R2 .
Mihaly Bencze
PP28746. If a, b, c, x, y, z > 0 then�a2 + b2 + c2
� �1x2 + 1
y2+ 1
z2
�+ 2(ab+bc+ca)(x+y+z)
xyz ≥
≥ 13 (a+ b+ c)2
�1x + 1
y + 1z
�2.
Mihaly Bencze
Proposed Problems 901
PP28747. If ak > 0 (k = 1, 2, ..., n) , then
P
cyclic
qa1a2a1+a2
!2
≤ n2
nPk=1
ak.
Mihaly Bencze
PP28748. In all acute triangle ABC holdsP sin2 A
A ≥ 4.
Mihaly Bencze
PP28749. In all triangle ABC holdsP�
ra
qr2b + rb + 1 + rb
pr2a + ra + 1
�2
r2c ≥≥ 2s2
�s2 − r2 − 4Rr + 2s2r (1 + 3r)
�.
Mihaly Bencze
PP28750. If a, b, c, d, e ∈ [1, 2] then�a2 + b2 + c2 + d2 − 3abcd
� �b2 + c2 + d2 + e2 − 3bcde
�·
·�c2 + d2 + e2 + a2 − 3cdea
� �d2 + e2 + a2 + b2 − 3deab
�·
·�e2 + a2 + b2 + c2 − 3eabc
�≤ 1.
Mihaly Bencze
PP28751. If x, y, z ∈ (0, 1) then
4096�1 + x2
� �1 + y2
� �1 + z2
� �9 + (x+ y + z)2
�3≥
≥ 729�4 + (x+ y)2
�2 �4 + (y + z)2
��4 + (z + x)2
�.
Mihaly Bencze
PP28752. If a, b, c ∈ (0, 1) and λ ≥ 2, then�43
�3λ �1 + aλ
� �1 + bλ
� �1 + cλ
� �3λ + (a+ b+ c)λ
�≥
≥�2λ + (a+ b)λ
��2λ + (b+ c)λ
��2λ + (c+ a)λ
�.
Mihaly Bencze
PP28753. Solve in Z the equation x3 + y3 = z�x2 − y2
�.
Mihaly Bencze
902 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP28754. In all triangle ABC holdsP
(mamb)λ ≤ 3
�(5R+2r)(s2−r2−4Rr)
12R
�λ
for all λ ∈ [0, 1] .
Mihaly Bencze
PP28755. Denote fk (n) the number of positive integers n such that kdivides
�2nn
�
1). ComputenP
p=1
1fk(p)
2). Compute∞Pp=1
1f2k(n)
when, k ∈ N, k ≥ 2.
Mihaly Bencze
PP28756. If ak ∈ [0, 1] (k = 1, 2, ..., n) , thennP
k=1
ak +nQ
k=1
ak ≥ 4n−1
P1≤i<j≤n
aiaj .
Mihaly Bencze
PP28757. If a, b > 0 and a3b3 − ab (a+ b) + 1 = 0 then�1
2a4+b+ 2
a4+a+b4
��1
2b4+a+ 2
b4+b+a4
�≤ 9
(2a+b)(a+2b) .
Mihaly Bencze
PP28758. If a, b, c > 0 and a+ b+ c = 32 then a2 + b2 + c2 + 32
9 abc ≥ 34 .
Mihaly Bencze
PP28759. Solve in Z the equation�x2+y2
xy+1
�2+
�y2+z2
1+yz
�2+
�z2+x2
1+zx
�2+ 1
xy + 1yz + 1
zx = 6.
Mihaly Bencze
PP28760. In all convex quadrilateral ABCD holdsP sinA+sinB(1+sinA) sinB ≤P 1
sinA .
Mihaly Bencze
Proposed Problems 903
PP28761. If x ∈�0, π
2n
�, then
nPk=1
sin kxk2 sin(k+1)x
> nn+1 .
Mihaly Bencze
PP28762. If a, b > 0 and a3b3 − ab (a+ b) + 1 = 0 then
9�2√16a2 + 9 +
√16b2 + 9− 1
��2√16b2 + 9 +
√16a2 + 9− 1
�≤
≤ 196 (2a+ b) (a+ 2b) .
Mihaly Bencze
PP28763. If a, b, c ∈ R and a2 + b2 + c2 ≤ 1 then |a|+ |b|+ |c| ≤ 4√3
3 +3abc.
Mihaly Bencze
PP28764. Denote Fk and Lk the kth Fibonacci respective Lucas numbers.Prove that1). 2FkFk+1
Fk+3+ 2FkFk+3
Fk+1+ 2Fk+1Fk+3
Fk≥ 2Fk+4
2). 2LkLk+1
Lk+3+ 2LkLk+3
Lk+1+ 2Lk+1Lk+3
Lk≥ 2Lk+4
Mihaly Bencze
PP28765. If a, b ≥ 1 then�2ab+1
a + 2a2
b
��2ab+1
b + 2b2
a
�≥ 4 (2a+ b) (a+ 2b) .
Mihaly Bencze
PP28766. If a, b > 0 and 2�a2 + b2
�+ 5ab+ 9 = 9 (a+ b) then
ab (a+ b) + 27a(a+2b) +
27b(b+2a) ≥ 18.
Mihaly Bencze
PP28767. If a, b > 0 then (a+b)2
4ab + a+2√ab
2a+b + b+2√ab
2b+a ≥ 33√a2b
3√
2a(a+b)2+ 3
3√ab2
3√
2b(a+b)2.
Mihaly Bencze
PP28768. If a, b > 0 then�ab
�2+�ba
�2+ 3ab
2(a+b)2≥ 23
4 .
Mihaly Bencze
PP28769. If a, b > 0 then ab +
ba + a(a+2b)
2a2+b2+ b(b+2a)
2b2+a2≥ 4.
Mihaly Bencze
904 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP28770. If a, b > 0 then a2
b + b2
a + 6a(a+2b)2a+b + 6b(b+2a)
2b+a ≥ 7 (a+ b) .
Mihaly Bencze
PP28771. If a, b > 0 and 2�a4 + b4
�+ 5a2b2 + 9 = 9
�a2 + b2
�then
a3
b + b3
a + (a+ b)2 + 23a
2b (2a+ b) + 23ab
2 (2b+ a) ≥ 6a2b2a+b +
6ab2
2b+a + 6.
Mihaly Bencze
PP28772. If a, b > 0 then ab +
ba + a+2
√ab
2a+b + b+2√ab
2b+a ≥ 4.
Mihaly Bencze
PP28773. If 0 < a1 < a2 < ... < an < an+1 < an+2 and
xn =�a2+a3a1+a2
�a2+a3+�a3+a4a2+a3
�a3+a4+ ...+
�an+1+an+2
an+an+1
�an+1+an+2
then
compute limn→∞
xn
n .
Mihaly Bencze
PP28774. If a, b > 0 then a3�b√a+1√ab
�+ b3
�a√b+1√ab
�+
√a+
√b
a3b3≥ 2
�ab+ a+b
ab
�.
Mihaly Bencze
PP28775. Determine all a, b ∈ N for which
2 (a+ b) = 3 +�√
a�√
2a��
+h√
b�√
2b�i
where [·] denote the integer.
Mihaly Bencze
PP28776. If a1 > 0 and (an − 1) (a1 + a2 + ...+ an−1) = 1 for all n ≥ 2
then compute∞Pk=1
1a21+a22+...+a2
k
.
Mihaly Bencze
PP28777. Let f : (0,+∞) → (0,+∞) a decreasing function for which exist
k > 0 such that limx→∞
f(x+k)f(x) = 1. Prove that lim
n→∞1n
nPp=1
f(x+k+p)f(x) = 1.
Mihaly Bencze
PP28778. If 3xn + 5xn = n+ 1 for all n ≥ 1, then compute limn→∞
xnxn+1
(lnn)2.
Mihaly Bencze
Proposed Problems 905
PP28779. Let be (xn)n≥1 a sequence for which the sequence (yn)n≥1 isconvergent, where yn = axn+2 + bxn+1 + cxn for all n ≥ 1. Determine alla, b, c ∈ R for which (xn)n≥1 is convergent.
Mihaly Bencze
PP28780. Prove that1R
−1
�√cos πx
2+�
cos 3πx5
�
dx
1+e−x ≤ 2.
Mihaly Bencze
PP28781. Prove that1R0
(x+ 1) esinπxdx ≥ 4e2π−1.
Mihaly Bencze
PP28782. Prove that 41R12
ln�1 + x2
�dx ≤ ln 2 +
1R−1
ln�1 + x2
�dx.
Mihaly Bencze
PP28783. Let be f : R → R where f (0) = 0 and f (x) =nQ
k=1
cos�
kekx−e−kx
�
for all x 6= 0. Prove that f is antiderivable.
Mihaly Bencze
PP28784. Solve in C the following system:
(xy)6 + 2√3 = 9 (yz)2
(yz)6 + 2√3 = 9 (zx)2
(zx)6 + 2√3 = 9 (xy)2
.
Mihaly Bencze
PP28785. Determine the function f : C → C wheref (z) = az3 + bz2 + cz + d (a, b, c, d ∈ C) such that |f (z)| = f (|z|) for allz ∈ C.
Mihaly Bencze
906 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP28786. If xk > 1 (k = 1, 2, ..., n) andnP
k=1
xn−1k = n (n− 1)n−1 then
Pcyclic
logx1
�xn−12 + xn−1
3 + ...+ xn−1n
�≥ n2.
Mihaly Bencze
PP28787. In all triangle ABC holdsP a2
b ≥ 2(2s2−3r2−12Rr)s .
Mihaly Bencze
PP28788. In all triangle ABC holdsP r2a
rb≥ 5(4R+r)2−12s2
4R+r .
Mihaly Bencze
PP28789. In all triangle ABC holdsP
tgA2 tg
B2
rtgA
2
tgC2
+P
tgA2
qtgB
2 tgC2 + 4r
s
PqtgA
2 tgB2 ≥ 12r
s�
�
tgA2tgB
2
+ 2.
Mihaly Bencze
PP28790. In all triangle ABC holds 2s
P√ab ≥ 11− s2+r2
2Rr .
Mihaly Bencze
PP28791. In all triangle ABC holdsP a2+b2
ac ≥ 2(5s2−3r2−12Rr)s2+r2+4Rr
.
Mihaly Bencze
PP28792. Solve the following system
[x] y2z + x {y} z = 108, 18[y] z2x+ y {z}x = 143, 635[z]x2y + z {x} y = 76, 534
,
where [·] and {·} denote the integer, respective the fractional part.
Mihaly Bencze
PP28793. In all triangle ABC holdsP a+b
b+c ≥ 2 + 6 3√2Rr3√s2+r2+2Rr
− 1s
P√ab.
Mihaly Bencze
PP28794. In all triangle ABC holdsP ra+rb
rb+rc≥ 2 + 6 3
pr4R − 2
�√rarb
4R+r .
Mihaly Bencze
Proposed Problems 907
PP28795. In all triangle ABC holds4(s2−r2−Rr)s2+r2+2Rr
+ s2+r2+4Rr2(s2−r2−4Rr)
≥ 4.
Mihaly Bencze
PP28796. If ak ∈ Z (k = 1, 2, ..., n) such thatnP
k=1
ak = 0 and�
nPk=1
ak�x2k−1
�= 0 for all x ∈ R, then a1 = a2 = ... = an = 0, where [·] and
{·} denote the integer, respective fractional part.
Mihaly Bencze and Marius Dragan
PP28797. In all triangle ABC holdsP �
ab
�2+P�
rarb
�2+ 15r
2R + 60Rrs2+r2+2Rr
≥ 272 .
Mihaly Bencze
PP28798. Compute limn→∞
n
��π4
�2 − n1R0
xnarctgxdx1+x2n
�.
Mihaly Bencze
PP28799. In all triangle ABC holdsP tg2 A
2+tg2 B
2
1+√3(tgA
2+tgB
2 )≥ 6
√3r+2s
4R+r+3s .
Mihaly Bencze
PP28800. Prove thatnP
k=1
�k 10√2018
< 17n(n+1)
40 where {·} denote the
fractional part.
Mihaly Bencze
PP28801. If r > 0 and λ ∈ [0, 1] thennP
k=1
({rk})λ ≤ n�n+12 max ({r} , 1− {r})
�λwhere {·} denote the fractional
part.
Mihaly Bencze
PP28802. Determine all ak ∈ N∗ (k = 1, 2, ..., n) for whichna1a2
o+n
a2a3
o+ ...+
nana1
o= 1 where {·} denote the fractional part.
Mihaly Bencze
908 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP28803. Prove thatnP
k=1
�n+
hn
1k + 1
k
i� 1k ∈ R\Q where [·] denote the
integer part.
Mihaly Bencze
PP28804. Denote Fk and Lk the kth Fibonacci respective Lucas numbers.Compute
1).∞Pk=1
12Fk
2).∞Pk=1
12Lk
Mihaly Bencze
PP28805. Prove thatnP
k=1
�log2
2n
k
�= 2n − 1 where [·] denote the integer
part.
Mihaly Bencze
PP28806. Determine all increasing functions f : N → N such that
n�
k=1
f(xk)
1+�
1≤i<j≤n
f(xi+xj)∈ N∗ for all xk ∈ N (k = 1, 2, ..., n) .
Mihaly Bencze
PP28807. In all triangle ABC holds
q(ax+ by + cz)2 + λ (a2 + b2 + c2) ≥ 4Area [ABC]
qP xyab + λ
P a2
b2for all
x, y, z,λ > 0 (A generalization of Weitzenbock’s inequality).
Mihaly Bencze
PP28808. In all triangle ABC holds
ax+ by + cz + λ�a2 + b2 + c2
�≥ 4Area [ABC]
�qP xyab + λ
qP a2
b2
�for all
x, y, z,λ > 0 (A generalization of Weitzenbock’s inequality).
Mihaly Bencze
Proposed Problems 909
PP28809. In all triangle ABC holds
ax+ by + cz + a2 + b2 + c2 ≥ 4Area [ABC]
�qP xyab +
qP a2
b2
�for all
x, y, z > 0 (A generalization of Weitzenbock’s inequality).
Mihaly Bencze
PP28810. In all triangle ABC holdsp(ax+ by + cz) (a2 + b2 + c2) ≥ 4Area [ABC] 4
r�P xyab
� �P a2
b2
�for all
x, y, z > 0 (A generalization of Weitzenbock’s inequality).
Mihaly Bencze
PP28811. Let ABC be a triangle, denote M the area of triangle formed by
ma,mb,mc. Prove that s2 − r2 − 4Rr ≥ 8M3
rP�
ma
mb
�2.
Mihaly Bencze
PP28812. Let ABC be a triangle, denote T the area of triangle formed by
cos A2 , cos
B2 , cos
C2 . Prove that 4R+ r ≥ 8T
sP�
cos A2
cos B2
�2
.
Mihaly Bencze
PP28813. In all triangle ABC holds
ax+ by + cz ≥ 4Area [ABC]q
xyab +
yzbc + zx
ca for all x, y, z > 0. (A
generalization of Weitzenbock’s inequality).
Mihaly Bencze
PP28814. In all triangle ABC holds
aλ + bλ + cλ ≥ 4Area [ABC]
qP(ab)λ−2 for all λ > 0 (A generalization of
Weitzenbock’s inequality).
Mihaly Bencze
PP28815. In all triangle ABC holds 3 + 2P
sin (4n+1)A2 sin A
2 ≥ 0 for alln ∈ N.
Mihaly Bencze
910 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP28816. In all triangle ABC holds
an + bn + cn ≥ 4Area [ABC]qP
(ab)n−2 for all n ∈ N,n ≥ 2. (A
generalization of Weitzenbock’s inequality).
Mihaly Bencze
PP28817. In all triangle ABC holds
1).P�
cosA+2 cosC2−cosB
�2cos4 A
2 ≤ (4R+r)2−s2
2R2
2).P�
cosA+2 cosC2−cosB
�2sin4 A
2 ≤ 8R2+r2−s2
2R2
Mihaly Bencze
PP28818. In all triangle ABC holdsP 3(cosA+2 cosC)2+16 cosB
(4+cos2 B)rarc≤ 4(4R+r)
s2r.
Mihaly Bencze
PP28819. In all triangle ABC holdsP 3(cosA+2 cosC)2+16 cosB
(4+cos2 B)(sinA+sinB)≤ 4R(5s2+r2+4Rr)
s(s2+r2+2Rr).
Mihaly Bencze
PP28820. In all triangle ABC holdsP (2−cosA)2
cosB+cosC ≥ 27(R+r)8R .
Mihaly Bencze
PP28821. In all triangle ABC holdsP (2−cosA)2
cosB+2 cosC ≥ 3(R+r)2R .
Mihaly Bencze
PP28822. In all triangle ABC holds
a2 + b2 + c2 ≥ 4Area [ABC]qP a2
b2≥ 4
√3Area [ABC] (A new refinement of
Weitzenbock’s inequality).
Mihaly Bencze
PP28823. In all triangle ABC holdsP 1
ab(s−a)(s−b) ≤�2R−r2sr2
�2.
Mihaly Bencze
Proposed Problems 911
PP28824. In all triangle ABC holdsP |(b−c)(c−a)|
ab ≤ (s2−r2−4Rr)(s2−3r2−12Rr)4s2r2
.
Mihaly Bencze
PP28825. In all triangle ABC holdsP 1
ab(b+c)(c+a) ≤�
s2−r2−Rrsr(s2+r2+2Rr)
�2.
Mihaly Bencze
PP28826. In all triangle ABC holdsP 1
ab(s−c) ≤(4R+r)2
4s3r2.
Mihaly Bencze
PP28827. In all triangle ABC holdsnP
k=1
(P
cos (2k + 1)A) ≤ 3n2 .
Mihaly Bencze
PP28828. In all triangle ABC holds 12 +P a2+b2
mamb≥ 20s2
�
mamb.
Mihaly Bencze
PP28829. In all triangle ABC holdsP√
bc cos2 A2 ≥ 3s
2 − 12
P�√a−
√b�2
.
Mihaly Bencze
PP28830. In all triangle ABC holdsP (a+b)(b+c)
ab ≤�s2+r2+4Rr
2sr
�2.
Mihaly Bencze
PP28831. In all triangle ABC holdsP a2
(a+b)(a+c) +P�
ab+c
�2≥ 8s2
5s2+r2+4Rr.
Mihaly Bencze
PP28832. In all triangle ABC holdsP 1√
(s−a)(a+b)(a+c)≤ s2−r2−Rr
2r√s(s2+r2+2Rr)
.
Mihaly Bencze
PP28833. In all acute triangle ABC holdsQ
(1− cosA) ≤ s2−(2R+r)2
4R2
�4R2
s2−(2R+r)2− 1
�3.
Mihaly Bencze
912 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP28834. In all triangle ABC holds
1).Q �
1− sin A2
�≤ r
4R
�4Rr − 1
�3
2).Q �
1− cos A2
�≤ s
4R
�4Rs − 1
�3
Mihaly Bencze
PP28835. In all triangle ABC holds
1).Q �
1 + sin A2
� 1
sin A2 ≥ 4
3
�1 + r
4R
� 4Rr
2).Q �
1 + cos A2
� 1
cos A2 ≥ 4
3
�1 + s
4R
� 4Rs
Mihaly Bencze
PP28836. In all triangle ABC holdsQ
(1− sinA) ≤ sr2R2
�2R2
sr − 1�3
.
Mihaly Bencze
PP28837. In all triangle ABC holdsQ
(1 + sinA)1
sinA ≥ 43
�1 + sr
2R
� 2R2
sr .
Mihaly Bencze
PP28838. In all acute triangle ABC holds
Q(1 + cosA)
1cosA ≥ 4
3
�1 + s2−(2R+r)2
4R2
� 4R2
s2−(2R+r)2 .
Mihaly Bencze
PP28839. If x > 0 then2(shx)
23
�
1+(shx)23
�
ch2x+ (shx)
43
1+sh4x+ 2(shx)
23
1+shx ≤ 3.
Mihaly Bencze
PP28840. Solve the following system:
h2x
x+{y}
i+h
2yy+{x}
i= 2h
x2
2x+{y}
i+h
y2
2y+{x}
i= 2
, when
[·] and {·} denote the integer, respective the fractional part.
Mihaly Bencze
PP28841. Compute limx→0
x�12 − arctgx−arcsinx
x3
�.
Mihaly Bencze
Proposed Problems 913
PP28842. If an = n (a1 + a2 + ...+ an−1) for all n ≥ 1 then determine alla1 ∈ N and all n ∈ N for which an have the last 500 decimal equals.
Mihaly Bencze
PP28843. If xk > 0 (k = 1, 2, ..., n) thenQ
cyclic
x21+x2
2x1+x2
≥nQ
k=1
xk.
Mihaly Bencze
PP28844. In all triangle ABC holds1).
Q �a2 + b2
�≥ 8s2Rr
�s2 + r2 + 2Rr
�
2).Q �
r2a + r2b�≥ 4s4Rr
3).Q �
sin4 A2 + sin4 B
2
�≥ r2((2R−r)(s2+r2−8Rr)−2Rr2)
512R5
4).Q �
cos4 A2 + cos4 B
2
�≥ s2((4R+r)3+s2(2R+r))
512R5
Mihaly Bencze
PP28845. Determine all n, k ∈ N for which fromzn1 − zk1 = zn2 − zk2 = zn3 − zk3 = zn4 − zk4 holds that two of there are equals,z1, z2, z3, z4 ∈ C.
Mihaly Bencze
PP28846. In all triangle ABC holdsPqsinA sinB
sin2 A−sinA sinB+sin2 B≥ s2+12Rr+3r2
s2−4Rr−r2.
Mihaly Bencze
PP28847. Solve in R the following system:
3x21−4x1 + x22 = 5x3 + 3x4
3x22−4x2 + x23 = 5x4 + 3x5
−−−−−−−−−−3x
21−4xn + x21 = 5x2 + 3x3
.
Mihaly Bencze
PP28848. If x1 > 0 and xn+1 =a1xn+b1[xn]+c1{xn}a2xn+b2[xn]+c2{xn} for all n ≥ 1 then prove
that the given sequence is convergent, and compute its limit, whereai, bi, ci > 0 (i = 1, 2) .
Mihaly Bencze
914 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP28849. In all triangle ABC holdsP s+rctgA
2�
tgA2+ctgA
2
≥ 6√sr.
Mihaly Bencze
PP28850. In all triangle ABC holds1).
P r+rara
√rra+rbrc
≥ 6s
2).P r+ha
ha
√rha+hbhc
≥ 3√2Rs
Mihaly Bencze
PP28851. If ak > 0 (k = 1, 2, ..., n) , thenP
cyclic
√19a21+2a1a2+4a22
4a2+a3≥ n.
Mihaly Bencze and Daniel Sitaru
PP28852. In all triangle ABC holds
1).Pp
4 + sin 2A+ 15 sin2A ≥ 1 + 4s+rR
2).P√
4 + sin 2A+ 15 cos2A ≥ 4 + 4r+sR
Mihaly Bencze
PP28853. Compute limn→∞
1R0
n√2016 + x ln (2017 + x) dx.
Mihaly Bencze
PP28854. If ak > 0 (k = 1, 2, ..., n) , thenP
cyclic
p76a21 + 230a1a2 + 178a22 ≥ 21
nPk=1
ak.
Mihaly Bencze
PP28855. In all triangle ABC holdsP �
tgA2 + tgB
2
�7 �tgC
2
�5 ≥ 128r2
9s2.
Mihaly Bencze
PP28856. In all triangle ABC holds
1).P�
1r − 1
ra
�71r2a
≥ 1289s4r5
2).P�
1r − 1
ha
�71h2a≥ 32R2
9s4r7
Mihaly Bencze
Proposed Problems 915
PP28857. In all triangle ABC holdsP 1
r5a(r5ar5c+r10b
+s2rr4br3c)
≤ (4R+r)2−2s2
(s2r)5.
Mihaly Bencze
PP28858. In all triangle ABC holdsP 1
h5a
�
h5ah
5c+h10
b+ 2s2r
Rh4bh3c
� ≤R3
�
(s2+r2+4Rr)2−16s2Rr
�
128s10r10.
Mihaly Bencze
PP28859. Prove that∞Pk=1
k2−1k8+5k4+1
≤ ζ(3)12 , where ζ denote the Riemann zeta
function.
Mihaly Bencze
PP28860. Prove thatnP
k=2
k3
k8+5k4+1≤ 3n2−n−2
48n(n+1) .
Mihaly Bencze
PP28861. If xk > 0 (k = 1, 2, ..., n) then compute maxnQ
k=1
xk
(k+xk)
Mihaly Bencze
PP28862. If ai > 0 (i = 1, 2, ..., n) , k ∈ N∗ then
Pcyclic
(a1+a2+...+an−1)k
(a1a2...an−1)k
n−1≥
(n−1)k+1
�
n�
i=1ai
�k+1
(�
n−1√a1a2...an−1)
k .
Mihaly Bencze
PP28863. If 1 < a < b thenbRa
x3dxx8+5x4+1
≤ 124 ln
(b−1)(a+1)(b+1)(a−1) .
Mihaly Bencze
PP28864. In all triangle ABC holds1). 16R ≤ 5r + s2
r
2). 4s√r(4R+r)
≤ 9s√r
(4R+r)√4R+r
+ (4R+r)√4R+r
s√r
Mihaly Bencze
916 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP28865. If an =nP
k=1
1√n2+k
then nn+1 ≤
nPk=1
a2k
k2≤ 2− 1
n .
Mihaly Bencze
PP28866. Determine all continuous functions f : [0, 1] → [0,+∞) for which
af (x) + f�x2
�+ f
�1− x2
�≤ a
1R0
f (x) dx, where a > 0.
Mihaly Bencze
PP28867. In all triangle ABC holds
maxnP 8h2
a+395h4
a+34h2a+80
;P 8r2a+39
5r4a+34r2a+80
o≤ 1
2r .
Mihaly Bencze
PP28868. In all tetrahedron ABCD holds
maxnP 8h2
a+395h4
a+34h2a+80
;P 8r2a+39
5r4a+34r2a+80
o≤ 1
2r .
Mihaly Bencze
PP28869. Solve in R the following system:
5x41 + 34x22 + 80 > 16x33 + 78x45x42 + 34x23 + 80 > 16x34 + 78x5−−−−−−−−−−−−−−5x4n + 34x21 + 80 > 16x32 + 78x3
.
Mihaly Bencze
PP28870. Prove thatnP
k=1
8k2+39(k+1)(5k4+34k2+80)
< n2n+1 .
Mihaly Bencze
PP28871. Determine all n ∈ N for which for all i, j ∈ {0, 1, ..., n}�ni
�divide�
nj
�or
�nj
�divide
�ni
�.
Tibor Jakab
Proposed Problems 917
PP28872. Solve in R the following system:
h3x2+1
3
i+h3y2+2
3
i=h2z2+1
2
ih3y2+1
3
i+h3z2+2
3
i=h2x2+1
2
ih3z2+1
3
i+h3x2+2
3
i=h2y2+1
2
i , when [·] denote the integer part.
Mihaly Bencze
PP28873. If xk ≥ 1 (k = 1, 2, ..., n) then
(x1 + x3)√x2 − 1 + (x2 + x4)
√x3 − 1 + ...+ (xn + x2)
√x1 − 1 ≤
≤ x1x2 + x2x3 + ...+ xnx1.
Mihaly Bencze
PP28874. Determine all ak > 0 (k = 1, 2, ..., n) if a1 = 1 and
6n (n+ 1)nP
k=1
kak = (2n+ 1) a2na2n+1.
Mihaly Bencze
PP28875. Computeh1k
2
i+ k
h2k
3
i+ k2
h3k
4
i+ ...+ kn−1
hnk
n+1
iwhen [·]
denote the integer part.
Mihaly Bencze
PP28876. If x, y, z ≥ 1 thenP 1+x2
(2+x2+y2)(2+x2+z2)≤
�
(1+xy)2
4(1+x2)(1+y2)(1+z2).
Mihaly Bencze
PP28877. If xk ≥ 1 thenP
cyclic
(1+x21)(1+x2
2)1+x1x2
≤ n+nP
k=1
x2k.
Mihaly Bencze
PP28878. If xk > 0 (k = 1, 2, ..., n) andnP
k=1
x2k = n then
Pcyclic
(1+x21)(1+x2
2)1+x1x2
≥ 2n.
Mihaly Bencze
918 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP28879. If xk > 0 (k = 1, 2, ..., n) andnP
k=1
x2k = n then
Pcyclic
(1+x21)(1+x2
2)...(1+x2r)
1+x1x2...xr≥ n2r−1, when r ∈ {2, 3, ..., n} .
Mihaly Bencze
PP28880. If bn =nP
k=1
2k−1h
k2
k+1
ithen compute
∞Pn=2
1bn.
Mihaly Bencze
PP28881. Solve in R the equation
h1
2017−x
i+h
11−2017x
i= 1
2017−[x] +1
1−[2017x] when [·] denote the integer part.
Mihaly Bencze
PP28882. Determine all a, b > 0 for which
�rha2
2
i+h(b+1)2
2
i�+
�rhb2
2
i+h(a+1)2
2
i�≥ 1 when [·] and {·} denote the
integer respective the fractional part.
Mihaly Bencze
PP28883. Let (bn)n≥1 be a geometrical progression for which b1 > 0, q > 1
when q is the ratio. Prove thatnP
k=1
bk+2−bk+1
(2k−1)(b2k−2−b1)≤ n
2n+1 .
Mihaly Bencze
PP28884. If ak > 0 (k = 1, 2, ..., n) such thatnP
k=1
a4k = 1 then
Pcyclic
√a1
1+a2 4√n≥ 1
2n .
Mihaly Bencze
PP28885. If a1 = 1 then determine all ak > 0 (k = 1, 2, ..., n) for whicha211 +
a222 + ...+ a2n
n =a31+a32+...+a3na1+a2+...+an
.
Mihaly Bencze
Proposed Problems 919
PP28886. If ak ≥ 1 (k = 1, 2, ..., n) andnP
k=1
ak = n+ 1 then
Pcyclic
pa1a2...an−1 (an − 1) ≤
r Pcyclic
a1a2...an−1.
Mihaly Bencze
PP28887. If an =�√
1�+�√
3�+�√
5�+ ...+
h√4n2 + 4n− 1
i, when [·]
denote the integer part, then compute∞Pk=1
1ak.
Mihaly Bencze
PP28888. If a1 =16 and an+1 =
n+1n+3
�an + 1
2
�for all n ≥ 1 then compute
nPk=1
[ak], when [·] denote the integer part.
Mihaly Bencze
PP28889. If Ak =2k+1Pp=0
hpk2 + 2p
iwhen [·] denote the integer part, then
compute∞Pk=1
1A2
k
.
Mihaly Bencze
PP28890. Determine all arithmetical progression (an)n≥1 and (bn)n≥1 forwhich (anbn)n≥1 is arithmetical progression too.
Mihaly Bencze
PP28891. Determine all geometrical progression (an)n≥1 and (bn)n≥1 forwhich (an + bn)n≥1 is geometrical progression too.
Mihaly Bencze
PP28892. If a, b, c > 0 thennP
k=1
�a
a2+k2(k+1)2+ b
b2+k2(k+1)2+ c
c2+k2(k+1)2
�≤ 3n
2(n+1) .
Mihaly Bencze
920 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP28893. In all triangle ABC holds1).
P aa2+rbrc
≤ 12r
2).P a
a2+hbhc≤ 1
2r
Mihaly Bencze
PP28894. Solve in R the equation
{x}+�x+ 1
n
+�x+ 1
n2
=
�n3x
+ 1
n + 1n2 when n ∈ N∗.
Mihaly Bencze
PP28895. In all triangle ABC holds
max���tgA
2 − tgB2
�� ;��tgB
2 − tgC2
�� ;��tgC
2 − tgA2
�� ≤ (4R+r)2
3s2.
Mihaly Bencze
PP28896. In all triangle ABC holdsP a
a2+(sin B2sin C
2 )4 ≤ 4R(2R−r)
r2.
Mihaly Bencze
PP28897. In all triangle ABC holdsP a
a2+(cos B2cos C
2 )4 ≤ 4R(4R+r)
s2.
Mihaly Bencze
PP28898. In all triangle ABC holdsP a
a2+(s−b)4(s−c)4≤ s2−2r2−8Rr
2s2r4.
Mihaly Bencze
PP28899. Denote B1 = {L1} , B2 = {L2, L3} , B3 = {L4, L5, L6, } ..., whereLk denote the kth Lucas number. Determine the sum of elements of the setBn.
Mihaly Bencze
PP28900. If a ∈ Q then the equation [x] {x} = a have an infinitely manysolutions in Q.
Mihaly Bencze
Proposed Problems 921
PP28901. If ak > 0 (k = 1, 2, ..., n) thenP
cyclic
a1a21+a2a3...an
≤n�
k=1
√ak
2
�
n�
k=1ak
.
Mihaly Bencze
PP28902. Denote A1 = {F1} , A2 = {F2, F3} , A3 = {F4, F5, F6} , ..., whenFk denote the kth Fibonacci number. Determine the sum of elements of setAn.
Mihaly Bencze
PP28903. If a1 = a+ 12 , a ∈ N∗ and an+1 = a2n − an + 3
4 for all n ≥ 1 thenprove that the number an+1 + an+2 is composite for all n ≥ 1.
Mihaly Bencze
PP28904. Let (an)n≥1 be an arithmetical progression with a1 > 0, r > 0and 4a1a2 < r, where r is the ratio. Denote
Sn =q1 + r2
a1a2a3+q
1 + r2
a2a3a4+ ...+
q1 + r2
anan+1an+2. Prove that
[Sn] = n for all n ≥ 3, when [·] denote the integer part.
Mihaly Bencze
PP28905. If a1 = 2, a2 = 6 and4an+2 = (n+ 3)
�4 +
√4an+1 + 1 +
√4an + 1
�for all n ≥ 1, then
nPk=1
1akak+2
= 118 − 1
3(n+1)(n+2)(n+3) .
Mihaly Bencze
PP28906. If ak ∈ [0, 1] (k = 1, 2, ..., n) , thenP
cyclic
a1a2+a33+3n−2
≤ 13 .
Mihaly Bencze
PP28907. Determine all p prime for which p2 + p+ 1 is prime andp2
n+ p2
n−1+ 1 is divisible by p2 + p+ 1 for all n ≥ 1.
Mihaly Bencze
922 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP28908. How many distinct divisors have the numbersnQ
k=1
�p2
k+ p2
k−1�
when p is a prime.
Mihaly Bencze
PP28909. If Sk =3n+1Pp=1
��p3
��kwhen [·] denote the integer part, then
compute S1, S2, S3, S4.
Mihaly Bencze
PP28910. Prove that3n+1Pk=1
��k3
��2 ≥ n2(3n+1)4 , where [·] denote the integer
part.
Mihaly Bencze
PP28911. If Sn =3n+1Pk=1
�k3
�when [·] denote the integer part, then
nPp=1
Sp =n(n+1)2
2 .
Mihaly Bencze
PP28912. Prove that 172n+ 172
n−1+ 1 is divisible by 307 for all n ≥ 1.
Mihaly Bencze
PP28913. Solve in Z the equation x2017
y+z + y2017
z+x + z2017
x+y = 32 .
Mihaly Bencze
PP28914. In all triangle ABC holds:
1).Pq
(s−a)(s−b)ac ≤ 3
2
2).Pr
cos A2cos B
2
(cos A2+cos B
2 )(cosB2+cos C
2 )≤ 3
2
Mihaly Bencze
PP28915. In all triangle ABC holdsP sin A
2
sin B2+(sin C
2 )3+7
≤ 13 .
Mihaly Bencze
Proposed Problems 923
PP28916. If dk =∞Pn=1
1(Ln)
k when Ln denote the nth Lucas number, then
compute∞Pk=1
(−1)kdkk! .
Mihaly Bencze
PP28917. Solve in Z the equation x3
y+z + y3
z+x + z3
x+y = −6.
Mihaly Bencze
PP28918. Let be xk =∞Pn=1
1pkn
where pn denote the nth prime. Compute
∞Pk=1
(−1)kxk
k! .
Mihaly Bencze
PP28919. Solve in Z the equation x2
y+z + y2
z+x + z2
x+y = −3.
Mihaly Bencze
PP28920. Compute∞R0
∞R0
∞R0
ln(x+y+z)dxdydz(ex+1)(ey+1)(ez+1) .
Mihaly Bencze
PP28921. If ck =∞Pn=1
1(Fn)
k where Fn denote the nth Fibonacci number,
then compute∞Pk=1
(−1)kCk
k! .
Mihaly Bencze
PP28922. Let be
1 + x(1−x)2
+ x2
(1−x)2(1−x2)2+ x3
(1−x)2(1−x2)2(1−x3)2+ ... =
∞�
n=0anxn
(1−x)2(1−x2)2(1−x3)2....
Compute∞Pn=0
1a20+a21+...+a2n
.
Mihaly Bencze
924 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP28923. Let be ak =∞Pn=1
1nk . Compute
∞Pk=1
(−1)kakk! .
Mihaly Bencze
PP28924. Let be bk =∞Pn=1
1(2n+1)k
. Compute∞Pk=1
(−1)kbkk! .
Mihaly Bencze
PP28925. Prove that∞Pk=1
arctg�6−
√5
3√5
�3+
√5
2
�n− 6+
√5
3√5
�3−
√5
2
�n�= π
4 .
Mihaly Bencze
PP28926. Prove that
����������
1 1 0 0 0 ... 013 1 2 0 0 ... 017
13 1 3 0 ... 0
− − − − − − −1
2n−11
2n−1−1... ... ... ... 1
����������
= n!1·3·7·...·(2n−1) .
Mihaly Bencze
PP28927. Prove that∞Pn=1
1n�
k=1(−1)n+k(nk)kn
= e− 1.
Mihaly Bencze
PP28928. Compute
1).∞Pk=1
Γ(sin2 kx)k!
2).∞Pk=1
Γ(cos2 kx)k! , where Γ denote the gamma function.
Mihaly Bencze
PP28929. Let A = (aij)1≤i,j≤n when aij =hij
i+hji
iwhere [·] denote the
integer part. Compute detA.
Mihaly Bencze
Proposed Problems 925
PP28930. Compute Ik =∞R0
e−x (lnx)k dx.
Mihaly Bencze
PP28931. Compute Vn =1R0
1R0
...1R0
x21+2x2
2+...+nx2n
nx1+(n−1)x2+...+xndx1dx2...dxn.
Mihaly Bencze
PP28932. Compute∞Pn=1
1p1+p2+...+pn
where pk denote the kth prime.
Mihaly Bencze
PP28933. Let be A = (aij)1≤i,j≤n where aij = iεj + jεi where εn = 1(ε 6= 1) . Compute detA.
Mihaly Bencze
PP28934. Let be A = (aij)1≤i,j≤n where aij =�i+ji
��i+jj
�. Compute detA.
Mihaly Bencze
PP28935. Compute
1).∞Pk=1
x2k sin(2k−1)x(2k+1)!
2).∞Pk=1
x2k cos(2k−1)x(2k+1)!
Mihaly Bencze
PP28936. Compute∞Pk=1
xk+yk
(x+y)k.
Mihaly Bencze
PP28937. Prove that
0 <
bRa
3xx2+x+1
dx
!2
−
bRa
3x sinxdxx2+x+1
!2
−
bRa
3x cosxdxx2+x+1
!2
≤ 112 .
Mihaly Bencze
926 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP28938. Compute1). 1− cosx
2 +B1cos 2x2! + ...+ (−1)nBn
cos 2nx(2n)! + ...
2). 1− sinx2 +B1
sin 2x2! + ...+ (−1)nBn
sin 2x(2n)! + ... when Bn denote the nth
Bernoulli number.
Mihaly Bencze
PP28939. Denote F and L the set of Fibonacci, respective Lucas numbers,and be x1 = 2, x2 = 8, xn+2 − 4xn+1 + xn = 0.
1). Compute∞Pn=1
1n2+y2n
where yn = {xn|n ∈ N} ∩ F
2). Compute∞Pn=1
1n2+z2n
where zn = {xn|n ∈ N} ∩ L
Mihaly Bencze
PP28940. Prove that∞Pk=1
arctg(7+4
√3)
k−2+(7−4√3)
k
3 = π2 .
Mihaly Bencze
PP28941. Solve in Z the equation y2 = 1 + 2x+ 3x2 + ...+ (n+ 1)xn.
Mihaly Bencze
PP28942. In all triangle ABC holds
1).P
sinAesinA ≥P cos A2 cos B−C
2 ecosA2cos B−C
2
2).P
cosAecosA ≥P sin A2 cos B−C
2 esinA2cos B−C
2
Mihaly Bencze
PP28943. If xk > 0 (k = 1, 2, ..., n) , thennP
k=1
xkexk ≥ P
cyclic
x1+x22 e
x1+x22 .
Mihaly Bencze
PP28944. In all triangle ABC holdsP 1sin2 A cos2 A
≤ 4 + 1sin2 A sin2 B sin2 C
+ 1cos2 A cos2 B cos2 C
.
Mihaly Bencze
PP28945. Prove that 1k + 2k + ...+ nk ≤ n− 1 + (n!)k , where k ∈ N .
Mihaly Bencze
Proposed Problems 927
PP28946. Prove that 2n ≤ n+ (n!)n+1
(1!2!...n!)2for all n ∈ N∗.
Mihaly Bencze
PP28947. If xk, yk ≥ 1 (k = 1, 2, ..., n) then�n− 1 +
nQk=1
x2k
��n− 1 +
nQk=1
y2k
�≥
�nP
k=1
xkyk
�2
.
Mihaly Bencze
PP28948. If ak > 0 (k = 1, 2, ..., n) , thennP
k=1
1ak+1 ≤ n− 1 + 1
1+n�
k=1ak
.
Mihaly Bencze
PP28949. If xk ≥ 1 (k = 1, 2, ..., n) , then
nx1 + (n− 1)x2 + ...+ 2xn−1 + xn ≤ n(n−1)2 + x1 + x1x2 + ...+ x1x2...xn.
Mihaly Bencze
PP28950. In all triangle ABC holds1).
P a2−ab+b2
(a2+b2)(2(a2+b2)−ab)≤ 1
12Rr
2).P r2a−rarb+r2
b
(r2a+r2b)(2(r2a+r2
b)−rarb)≤ 4R+r
6s2r
Mihaly Bencze
PP28951. Prove thatnP
k=1
k2+k+1(2k2+2k+1)(3k2+3k+2)
≤ n6(n+1) .
Mihaly Bencze
PP28952. In all triangle ABC holds
1).P sin4 A
2−sin2 A
2sin2 B
2+sin4 B
2
(sin4 A2+sin4 B
2 )(2(sin4 A
2+sin4 B
2 )−sin2 A2sin2 B
2 )≤ 4R(2R−r)
3r2
2).P cos4 A
2−cos2 A
2cos2 B
2+cos4 B
2
(cos4 A2+cos4 B
2 )(2(cos4A2+cos4 B
2 )−cos2 A2cos2 B
2 )≤ 4R(4R+r)
6s2
Mihaly Bencze
928 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP28953. If xk ≥ 1 (k = 1, 2, ..., n) , thennQ
k=1
(x1 + x2 + ...+ xk) ≤nQ
k=1
(k − 1 + x1x2...xn) .
Mihaly Bencze
PP28954. If xk > 0 (k = 1, 2, ..., n) , then 1 +nP
k=1
sh2xk ≤nQ
k=1
ch2xk.
Mihaly Bencze
PP28955. Prove thatnP
k=1
�1k2
+ 1(k+1)2
+ (2k+1)4
2k2(k+1)2(2k2+2k+1)
�≥ 6n
n+1 .
Mihaly Bencze
PP28956. If xk > 0 (k = 1, 2, ..., n) , then
13
nPk=1
x2k ≥ Pcyclic
(x21−x1x2+x2
2)x1x2
(x21+x2
2)(2(x21+x2
2)−x1x2).
Mihaly Bencze
PP28957. In all tetrahedron ABCD holds
1).�
1ha
+ 1hc
�hd�
1hb
+ 1hd
�hc�
rhc
+ rhd
�hc+hd ≤�
1hc
� 1hd
�1hd
� 1hc
2).�
1ra
+ 1rc
�rd�
1rb
+ 1rd
�rc �r2rc
+ r2rd
�rc+rd ≤�
1rc
�rd�
1rd
�rc
Mihaly Bencze
PP28958. If a, b, c > 0 thenP a+b
(b+c)(c+a) ≥9
2(a+b+c) .
Mihaly Bencze
PP28959. If a, b, c > 0 and λ > 0 thenP a+λb+c
(b+λc+a)(c+λa+b) ≥9
(λ+2)(a+b+c) .
Mihaly Bencze
PP28960. If a, b, c > 0 then�1 + a
c
�2c �1 + b
a
�2a �1 + c
b
�2b ≤�2a+b+ca+c
�a+c �2b+c+ab+a
�b+a �2c+a+bc+b
�c+b.
Mihaly Bencze
Proposed Problems 929
PP28961. In all triangle ABC holds
1).�1 + rc
ra
��1 + 1
xrb
�xrc � r(xrc+1)rc(xr+1)
�xrc≤ 1
2).�1 + hc
ha
��1 + 1
xhb
�xhc�r(xhc+1)hc(xr+1)
�xhc
≤ 1 for all x > 0
Mihaly Bencze
PP28962. If a, b, c > 0 then�1 + a
c
�c �a+2b+ca+b+c
�a+b+c≤
�2(a+b+c)a+b+2c
�a+b+2c
and his permutations.
Mihaly Bencze
PP28963. If n ∈ N ∗ then�1 + 1
n
�n �n+2n+3
�n+3≤ 4
�2n+32n+5
�2n+5.
Mihaly Bencze
PP28964. If a, b > 0 and x ∈ R then�1 + a
sin2 x
�sin2 x �1 + b
cos2 x
�cos2 x ≤ a+ b+ 1.
Mihaly Bencze
PP28965. Prove that
1).�1 + Fk
Fk+2
�Fk+2�1 +
Fk+1
Fk+3
�Fk+3 ≤�Fk+2+Fk+4
Fk+1+Fk+3
�Fk+4
2).�1 + Lk
Lk+2
�Lk+2�1 +
Lk+1
Lk+3
�Lk+3 ≤�Lk+2+Lk+4
Lk+1+Lk+3
�Lk+4
for all k ∈ N∗
Mihaly Bencze
PP28966. In all tangential quadrilateral ABCD with radius R andsemiperimeter s holds
�1 + tgC
2 ctgA2
�ctgC2�1 + tgB
2 ctgD2
�ctgD2 ≤
�s
R(ctgC2+ctgB
2 )
�ctgC2+ctgD
2
.
Mihaly Bencze
930 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP28967. In all triangle ABC holds
1).P�
(2+ctg2 A2 ) ln(2+ctg2 A
2 )ln(1+ctg2 A
2 )
� 1
2+ctg2 A2 ≥ 18R
8R−r
2).P�
(2+tg2 A2 ) ln(2+tg2 A
2 )ln(1+ctg2 A
2 )
� 1
2+tg2 A2 ≥ 18R
10R+r
Mihaly Bencze
PP28968. In all acute triangle ABC holds
1).P�
(2+tg2A) ln(2+tg2A)ln(1+tg2A)
� 12+tg2A
≥ 18R2
9R2−s2+(2R+r)2
2).P�
(2+ctg2A) ln(2+ctg2A)ln(1+ctg2A)
� 12+ctg2A
≥ 18R2
6R2+s2−r(4R+r)
Mihaly Bencze
PP28969. In all triangle ABC holds 1R√Rr
≤P sinA
ra√
a(s−a)≤ 1
2r√Rr
(A
refinement of Euler’s R ≥ 2r inequality).
Mihaly Bencze
PP28970. In all triangle ABC holds s4r2
≥P m2a
s−a ≥ s2Rr (A refinement of
Euler’s R ≥ 2r inequality).
Mihaly Bencze
PP28971. In all triangle ABC holds 3R4sr2
≥P 1(b+c)(s−a) ≥
32sr (A
refinement of Euler’s R ≥ 2r inequality).
Mihaly Bencze
PP28972. In all triange ABC holds 9R2
r2(s2+(4r+r)2)≥P 1
m2a≥ 18R
r(s2+(4R+r)2)(A refinement of Euler’s R ≥ 2r inequality).
Mihaly Bencze
Proposed Problems 931
PP28973. In all triangle ABC holds 4√RrP
ra
qs−aa ≥ 4
√3Area [ABC]
(Weitzenbock’s type inequality).
Mihaly Bencze
PP28974. In all triangle ABC holds 2sr2
R+r
P √b2+bc+c2
s−a ≥ 4√3Area [ABC]
(Weitzenbock’s type inequality).
Mihaly Bencze
PP28975. In all triangle ABC holds 13r2
≥P 1
wawb≥ 2
3Rr (A refinement ofEuler’s R ≥ 2r inequality).
Mihaly Bencze
PP28976. In all acute triangle ABC holds
2P a2 sin A
2
cosA cos B−C2
≥ 4√3Area [ABC] (Weitzenbock’s type inequality).
Mihaly Bencze
PP28977. Prove thatnP
k=1
e−xk+1−e−xk
x dx = nγn+1 where γ is the
Euler-Mascheroni constant.
Mihaly Bencze
PP28978. Prove thatnP
k=1
e−x2k+1−e−x2k−1
x dx = nγ2n+1 where γ is the
Euler-Mascheroni constant.
Mihaly Bencze
PP28979. In all triangle ABC holdsP
AB∞R0
e−xA−e−xB
x dx = 0.
Mihaly Bencze
PP28980. In all triangle ABC holdsP
cos2A sin2A∞R0
e−xcos2 A−e−xsin
2 A
x dx =((2R+r)2−R2−s2)γ
R2 , where γ is the
Euler-Mascheroni constant.
Mihaly Bencze
932 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP28981. If xk > 0 (k = 1, 2, ..., n) , thennP
k=1
∞R0
e−tch2x−e−tsh
2t
t dt ≥ n2−γn�
k=1
sh2xkch2xk
where γ is the Euler-Mascheroni
constant.
Mihaly Bencze
PP28982. Prove that, if 0 < a ≤ b thenbRa
�∞R0
e−xch2t−e−xsh
2t
x dx
�dt = 2γsh2(a−b)
sh2ash2b where γ is the Euler-Mascheroni
constant
Mihaly Bencze
PP28983. In all triangle ABC holds
1γ
P ∞R0
e−xsin2 A−e−xcos
2 A
x dx =�
s2+r2−4R2
s2−(2R+r)2
�2− 8R(R+r)
s2−(2R+r)2−�s2+r2+Rr
2sr
�2+ 4R
r
where γ is the Euler-Mascheroni constant.
Mihaly Bencze
PP28984. Compute∞P
k1=1
∞Pk2=1
∞Pk3=1
1(k21+k22)(k
22+k23)(k
23+k21)
.
Mihaly Bencze
PP28985. Compute∞R0
sinx(1−(sinx)2n+2)cos2 x
dx where n ∈ N.
Mihaly Bencze
PP28986. Compute∞R0
cosx(1−(cosx)2n+2)dxsin2 x
where n ∈ N.
Mihaly Bencze
PP28987. Compute
π2R0
tgx ln�1+sin 2x1−sin 2x
�dx.
Mihaly Bencze
Proposed Problems 933
PP28988. Compute∞Pn=1
Hn
ζ (2k)−
nPp=1
1p2k
!, where ζ denote the
Riemann zeta function and Hn =nP
i=1
1i and k ∈ N∗.
Mihaly Bencze
PP28989. Compute
π2R0
π2R0
ln(sinx) ln(cos y) ln(1−cosx) ln(1−sin y)dxdysinx cos y .
Mihaly Bencze
PP28990. Let be f (x) =nP
k=1
(k (k + 1))x − n(n+1)(n+2)3 . Compute
limn→∞
1n3
n(n+1)(n+2)3R0
f−1 (x) dx.
Mihaly Bencze
PP28991. Prove that if 0 < a ≤ b < 1 thenbRa
(√x+
√1−x) ln(x−x2)1+
√2x
dx =√2 ln bb(1−a)1−a
aa(1−b)1−b .
Mihaly Bencze
PP28992. Prove that if 0 < a ≤ b < 1 thenbRa
(√x+
√1−x)(sinπx+sinπ(1−x))dx
1+√2x
= 4√2
π sin (b−a)π2 cos π(b+a−1)
2 .
Mihaly Bencze
PP28993. If xk > 0 (k = 1, 2, ..., n) andnP
k=1
xk = n thennQ
k=1
(1+xk)1+xk
xxkk
≤ 4n.
Mihaly Bencze
PP28994. If x, y, z > 0 then 64(x+1)x+1(y+1)y+1(z+1)z+1(x+y+z+3)x+y+z+3
xxyyzz(x+y+z)x+y+z ≤
≤ 27(x+y+2)x+y+2(y+z+2)y+z+2(z+x+2)z+x+2
(x+y)x+y(y+z)y+z(z+x)z+x .
Mihaly Bencze
934 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP28995. In all triangle ABC holds
1). (a+1)a+1(b+1)b+1(c+1)c+1
aabbcc≤
�1 + 2s
3
�3 �1 + 3
2s
�2s
2). (ra+1)ra+1(rb+1)rb+1(rc+1)rc+1
rraa rrbbrrcc
≤�1 + 4R+r
3
�3 �1 + 3
4R+r
�4R+r
Mihaly Bencze
PP28996. If xk > 0 (k = 1, 2, ..., n) then
nQk=1
(xk + 1)�1 + 1
xk
�xk ≤�1 + 1
n
nPk=1
xk
�n1 + n
n�
k=1
xk
n�
k=1
xk
.
Mihaly Bencze
PP28997. If x, y, z > 0 and x+ y + z = 3 then4096(x+1)x+1(y+1)y+1(z+1)z+1
xxyyzz ≤ (5−x)5−x(5−y)5−y(5−z)5−z
(3−x)3−x(3−y)3−y(3−z)3−z .
Mihaly Bencze
PP28998. If x ∈�0, π2
�then 3 sin2 x cos2 x+ (tgx)6 cos
2 x + (ctgx)6 sin2 x ≥ 1.
Mihaly Bencze
PP28999. Denote Fk and Lk the kth Fibonacci respective Lucas numbers.Prove that1). (Fk+2)
Fk+2 ≥ (Fk)Fk (Fk+1)
Fk+1
1). (Lk+2)Lk+2 ≥ (Lk)
Lk (Lk+1)Lk+1 for all k ∈ N∗
Mihaly Bencze
PP29000. In all triangle ABC holds (n+ 1)P 1
cos A2
≥ (2n+ 1)√3 + 4R+ r
for all n ∈ N∗.
Mihaly Bencze
PP29001. In all triangle ABC holds (n+ 1)P
r4a ≥ s2r (12R+ (9n+ 7) r)for all n ∈ N∗.
Mihaly Bencze
Proposed Problems 935
PP29002. If a, b, c ∈ (0, 1) ∪ (1,+∞) thenP (an+1−1)(2a−b−c)
(a−1)(b+c) ≥ 0.
Mihaly Bencze
PP29003. If a, b, c > 0 thennP
k=2
�P a(b+c)2+(k2+k−4)a2
�≥ 3(n−1)
2(n+1) .
Mihaly Bencze
PP29004. If a, b, c > 0 then 3√3P
a2 ≥�a2 + b2 − c2
�+ 4
Pab.
Mihaly Bencze
PP29005. If a, b, c > 0 then 9�P
a2�2 ≥ 3
P �a2 + b2 − c2
�2+ 8 (
Pab)2 .
Mihaly Bencze
PP29006. In all triangle ABC holdsP a3
b+c−a ≥ 2(s2−r2−4Rr)s2+r2+4Rr
.
Mihaly Bencze
PP29007. In all triangle ABC holdsP (s−b)(s−c)
b2c2(s−a)≥
√3
8R3 .
Mihaly Bencze
PP29008. Determine all λ ≥ 0 for which in all triangle ABC holdsP a3
b+c−a ≥≥ 4√3sr + λ
P(a− b)2 .
Mihaly Bencze
PP29009. In all triangle ABC holds1
2sr
Pa3 (s− b) (s− c) ≥ 4
√3Area [ABC] (Weitzenbock’s type inequality).
Mihaly Bencze
PP29010. In all triangle ABC holds
1). rλ2P�
1ra
� 3λ2+ 6
sλ≤ r
λ−22P�
1ra
� 3λ−22
2). rλ2P�
1ha
� 3λ2+ 6
sλ≤ r
λ−22P�
1ra
� 3λ−22
for all λ > 0.
Mihaly Bencze
936 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP29011. Prove thatnQ
k=2
�(k+1) ln(k+1)
ln k
� 1k+1 ≥ 2
n+1 .
Mihaly Bencze
PP29012. If 1 < a ≤ b then ln b+1a+1 +
bRa
�(x+1) ln(x+1)
lnx
� 1x+1
dx ≥ b− a.
Mihaly Bencze
PP29013. If xk > 1 (k = 1, 2, ..., n) then
nPk=1
�(x2
k+1) ln(x2
k+1)
2 lnxk
� 1
x2k+1
≥
�
n�
k=1xk
�2
n+n�
k=1
x2k
.
Mihaly Bencze
PP29014. If xk > 1 (k = 1, 2, ..., n) , then
nPk=1
�(xk+1) ln(xk+1)
lnxk
� 2xk+1 ≥
�
n�
k=1xk
�2
n+2n�
k=1xk+
n�
k=1x2k
.
Mihaly Bencze
PP29015. If xk > 0 (k = 1, 2, ..., n) , thennP
k=1
sh2xk�
cos
�
π
sh2xk+ch2xk
�
≥ 1n
�nP
k=1
shxk
�2
+ 1n
�nP
k=1
chxk
�2
.
Mihaly Bencze
PP29016. Prove thatnP
k=1
1cos π
2(2k+1)≥ n(4n+5)
4(n+1) .
Mihaly Bencze
PP29017. In all triangle ABC holds
1).Qr
cos�π2
�a−ba+b
��≤ 16Rr
s2+r2+2Rr
2).Qq
cos�π2
�a−bc
��≤ 2r
R 3).Qr
cos�π2
�ra−rbra+rb
��≤ 2r
R
Mihaly Bencze
Proposed Problems 937
PP29018. If xk ∈�0, π2
�(k = 1, 2, ..., n) , then
nPk=1
1�
cos(π2cos 2xk)
≥ n
sin
�
2n
n�
k=1xk
� .
Mihaly Bencze
PP29019. If x, y ∈ R then
2 sin (x+ y) cos (x− y) ≥qcos
�π2 cos 2x
�+qcos
�π2 cos 2y
�.
Mihaly Bencze
PP29020. Prove that sin 2x ≥qcos
�π2 cos 2x
�for all x ∈ R.
Mihaly Bencze
PP29021. If a, b, c > 0 thenQr
cos�π2
�a−ba+b
��≤ 8abc
(a+b)(b+c)(c+a) .
Mihaly Bencze
PP29022. In all triangle ABC holdsPq
cos�π2 cos 2A
�≤ 2sr
R2 .
Mihaly Bencze
PP29023. If a, b, c > 0 thenQ
cos�π2
�a−c
a+2b+c
��≤
�8(a+b)(b+c)(c+a)
(a+2b+c)(b+2c+a)(c+2a+b)
�2.
Mihaly Bencze
PP29024. In all triangle ABC holdsP 1
a2 cos(π2 (
b−cb+c ))
≥ s(s2+r2−6Rr)16sR2r2
≥ 12Rr .
Mihaly Bencze
PP29025. In all triangle ABC holds
1).P
(a+ b) cos�π2
�a−ba+b
��≤
2�
(s2+r2+4Rr)2+8s2Rr
�
sRr
2).P
(ra + rb) cos�π2
�ra−rbra+rb
��≤ s2+r(4R+r)
R
3).P
(sinA+ sinB) cos�π2
�sinA−sinBsinA+sinB
��≤ (s2+r2+4Rr)
2−8s2Rr
sR(s2+r2+2Rr)
938 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
4).P
a cos�π2
�b−ca
��≤ r(s2+(4R+r)2)
sR
Mihaly Bencze
PP29026. Prove thatnP
k=1
k+1k2
cos�π2
�k−1k+1
��≤ 4n
n+1 .
Mihaly Bencze
PP29027. Prove thatnQ
k=1
rk cos
�π2
�k−1k+1
��≤ 2n
n+1 .
Mihaly Bencze
PP29028. If x, y, z ∈ [0, 1] then(z − y + 1) sinx+ (x− z + 1) sin y + (y − x+ 1) sin z ≤≤ sin ((x− y) z + y) + sin ((y − z)x+ z) + sin ((z − x) y + x) .
Mihaly Bencze
PP29029. In all triangle ABC holds
1). s2−r(4R+r)2R2 cosx+
�1− s2−(2R+r)2
2R2
�cos y ≤P cos
�x sin2A+ y cos2A
�
2). s2−r(4R+r)2R2 sinx+
�1− s2−(2R+r)2
2R2
�sin y ≤
Psin
�x sin2A+ y cos2A
�for
all x, y ∈�0, π2
�
Mihaly Bencze
PP29030. If x, y ∈�0, π2
�and λ ∈ [0, 1] then
λ sinx+(1−λ) sin ycos(λx+(1−λ)y) ≤ sin(λx+(1−λ)y)
λ cosx+(1−λ) cos y .
Mihaly Bencze
PP29031. If x, y, z ∈ [0, 1] then(z − y + 1) cosx+ (x− z + 1) cos y + (y − x+ 1) cos z ≤≤ cos ((x− y) z + y) + cos ((y − z)x+ z) + cos ((z − x) y + x) .
Mihaly Bencze
PP29032. If x, y ∈�0, π2
�then
nPk=1
cos
�x+(k2+k−1)y
k(k+1)
�≥ n
n+1 (cosx+ n cos y) .
Mihaly Bencze
Proposed Problems 939
PP29033. If x, y ∈�0, π2
�then
nPk=1
sin
�x+(k2+k−1)y
k(k+1)
�≥ n
n+1 (sinx+ n sin y) .
Mihaly Bencze
PP29034. If λ ∈ [0, 1] and x, y ∈�0, π2
�then
λ (1− λ) sin (x+ y) + λ2 sinx cosx+ (1− λ)2 sin y cos y ≤≤ 1
2 sin (2λx+ 2 (1− λ) y) .
Mihaly Bencze
PP29035. If λ ∈ [0, 1] and x1, y1, x2, y2 ∈�0, π2
�then
λ cos x1+x22 cos x1−x2
2 + (1− λ) cos y1+y22 cos y1−y2
2 ≤≤ cos λ(x1+x2)+(1−λ)(y1+y2)
2 cos λ(x1−x2)+(1−λ)(y1−y2)2 .
Mihaly Bencze
PP29036. If x, y ∈�0, π2
�and t ∈ R then
(cosx− cos y)2 sin2 t cos2 t+ cosx cos y ≤≤ cos
�sin2 tx+ cos2 ty
�cos
�cos2 tx+ sin2 ty
�.
Mihaly Bencze
PP29037. If t ∈ R and x, y ∈�0, π2
�then
sinx+ sin y ≤ sin�x sin2 t+ y cos2 t
�+ sin
�x cos2 t+ y sin2 t
�.
Mihaly Bencze
PP29038. If t ∈ R and x, y ∈�0, π2
�then
cosx+ cos y ≤ cos�x sin2 t+ y cos2 t
�+ cos
�x cos2 t+ y sin2 t
�.
Mihaly Bencze
PP29039. In all triangle ABC holds
1). 2 (cosx+ cos y) ≤P cos�
rrax+
�1− r
ra
�y�+P
cos��
1− rra
�x+ r
ray�
2). 2 (cosx+ cos y) ≤P cos�
rhax+
�1− r
ha
�y�+P
cos��
1− rha
�x+ r
hay�
for all x,y∈�0, π2
�.
Mihaly Bencze
940 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP29040. If ak > 0 (k = 1, 2, ..., n) , then�a1a2
+ 1�a2 �a2
a3+ 1
�a3· ... ·
�ana1
+ 1�a1
≤ 2a1+a2+...+an .
Mihaly Bencze
PP29041. If x, y ∈ R then�1 + sin2 x
sin2 y
�sin2 y �1 + cos2 x
cos2 y
�cos2 y≤ 2.
Mihaly Bencze
PP29042. In triangle ABC holds A ≥ B ≥ C. Prove thatc sinA+ b sinB + a sinC ≤ 2s2
3R .
Mihaly Bencze
PP29043. In all triangle ABC holdsP ra(λs−a)
s−a ≥ (4R+r)(4(λ−1)R+(λ+2)r)3r for
all λ ≥ 1.
Mihaly Bencze
PP29044. In all acute triangle ABC holds�s2 − r2 − 4Rr
� �s2 − (2R+ r)2
�+ 45
�s2 − (2R+ r)2
�2+ 16s2r2 ≤
≤ 2�s2 − r2 − 4Rr
�2.
Mihaly Bencze
PP29045. If x, y, z > 0 then in triangle ABC holds
xOI + yGI + zGO ≤√
x2+y2+z2
3
�18R2 − 26Rr + 7r2 − s2
�.
Mihaly Bencze
PP29046. Let A1B1C1 be the Morley triangle of ABC. Denote Rm and Rthe circumradii of triangles A1B1C1 and ABC. Prove thatRm
R ≤ 8√3
3
Qsin A+B
6 ≤ 8√3
3
Qsin π+A
12 ≤ 8√3
3
Qsin 3π−A
24 ≤ 8√3
3
Qsin 5π+A
48 ≤... ≤ 8
√3
3
�sin π
9
�3.
Mihaly Bencze and Mehmet Sahin
PP29047. In all triangle ABC holds nra+krb+prcn+k+p ≥ nrama+krbmb+prcmc
nma+kmb+pmcfor all
n, k, p ∈ N∗.
Mihaly Bencze
Proposed Problems 941
PP29048. Let ABC be an acute triangle, denote A1B1C1 his Morleytriangle, denote Rm and R the circumradii of triangles A1B1C1 and ABC.
Prove that 64√3ABC
81π3 < Rm
R < 8√3ABC81 .
Mihaly Bencze
PP29049. In all triangle ABC holds1).
P(ma +mb)
3 > 27s4
�s2 − 3r2 − 6Rr
�
2).P
(ma +mb)2 (mb +mc)
2 > 8116
��s2 + r2 + 4Rr
�2 − 16s2Rr�
3).P 1
(ma+mb)2 < 4
9
��s2+r2+4Rr
4sRr
�2− 1
Rr
�
4).P 1
(ma+mb)2(mb+mc)
2 <81(s2−r2−4Rr)
128s2R2r2
Mihaly Bencze
PP29050. In all triangle ABC holdsP ra
rb+rc≥ (4R+r)2
2s2.
Mihaly Bencze
PP29051. Let Ra, Rb, Rc be the circumradius of triangles BOC, COA,
AOB, in triangle ABC. Prove thatP Ra
R+2Ra= 1
4 +�4R+r2s
�2.
Mihaly Bencze
PP29052. In all triangle ABC holds
1).P�
1 + rra
� rrb ≤ 4
�1 + R
r
�
2).P�
1 + 1ha
� rhb ≤ 13
4 + r(4R+r)4s2
Mihaly Bencze
PP29053. In all triangle ABC holdsP�
rarb+rc
�λ≤ 1
3λ−1
�R3r − 1
6
�λfor all
λ ∈ [0, 1] .
Mihaly Bencze
PP29054. In all acute triangle ABC holdsPq
(1 + sinA)cosA (1 + cosA)sinA ≤ 3 + srR2 .
Mihaly Bencze
942 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP29055. In all triangle ABC holdsAI ·BI · CI = 2R2 (ha − 2r) (hb − 2r) (hc − 2r) .
Mihaly Bencze
PP29056. In all triangle ABC holdsP
(1 + sinA)sinB ≤ 3 + s2+r2+4Rr4R2 .
Mihaly Bencze
PP29057. In all acute triangle ABC holdsP(1 + cosA)cosB ≤ 3 + s2+r2−4R2
R2 .
Mihaly Bencze
PP29058. In all acute triangle ABC holdsP(1 + sinA)sinA (1 + cosA)cosA ≤ 6 + 2sr
R2 .
Mihaly Bencze
PP29059. In all triangle ABC holdsPq
2�1 + cos B
2 cos C2 − cos2 A
2
�+ r
2R ≤ 2P
cos A2 .
Mihaly Bencze
PP29060. In all triangle ABC holdsP ma(mb+mc)
mb+mc−ma≥ 2
Pma.
Mihaly Bencze
PP29061. In all triangle ABC holdsP
m2am
2b ≥ 9s2r2 + 3
2
�P ��a2 − b2���2 .
Mihaly Bencze
PP29062. If ak > 0 (k = 1, 2, ..., n) , thennQ
k=1
a2k+ak+1
ak(ak+1) ≥ 1 + 1n�
k=1ak
+ 1n�
k=1(ak+1)
.
Mihaly Bencze
PP29063. In all triangle ABC holdsP (cos A
2+cos B
2−cos C
2 )3
cos C2
≥ 4R+r2R .
Mihaly Bencze
Proposed Problems 943
PP29064. If ak > 0 (k = 1, 2, ..., n) , thenP
cyclic
a21√2a2a3+
√a22+a23
≥ 12√2
nPk=1
ak.
Mihaly Bencze
PP29065. If ak > 0 (k = 1, 2, ..., n) , thenPcyclic
a21
(a2+a3)2�√
2a2a3+√
a22+a23
� ≥ n2
8√2
n�
k=1ak
.
Mihaly Bencze
PP29066. Prove thatnP
k=1
1k(
√2k+
√k2+1)
≥ n√2(n+1)
.
Mihaly Bencze
PP29067. If ak > 0 (k = 1, 2, ..., n) thenP
cyclic
1√2a1a2+
√a21+a22
≥ n2
2n�
k=1ak
.
Mihaly Bencze
PP29068. In all triangle ABC holds:
1).Q�√
ab+q
a2+b2
2
�≤ 2s
�s2 + r2 + 2Rr
�
2).Q�p
(s− a) (s− b) +
q(s−a)2+(s−b)2
2
�≤ 4sRr
3).Q�
√rarb +
qr2a+r2
b
2
�≤ 4s2R
4).Q�√
hahb +
qh2a+h2
b
2
�≤ s2r(s2+r2+2Rr)
R2
5).Q�
sin A2 sin B
2 +q
12
�sin4 A
2 + sin4 B2
��≤ (2R−r)(s2+r2−8Rr)−2Rr2
32R3
6).Q�
cos A2 cos B
2 +q
12
�cos4 A
2 + cos4 B2
��≤ (4R+r)3+s2(2R+r)
32R3
Mihaly Bencze
PP29069. Denote Fk and Lk the kth Fibonacci respective Lucas numbers.Prove that
1).nP
k=1
�Fk −
√FkLk + Lk
�2 ≥ Fn+2+Ln+2
2 − 2
944 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
2).nP
k=1
�F 2k − FkLk + L2
k
�2 ≥ FnFn+1+LnLn+1
2 − 1
Mihaly Bencze
PP29070. If x ∈�0, π2
�then
1). sinx+ cosx ≥ 1√2+√sinx cosx
2). 1 ≥ sinx cosxq
12 − sin2 x cos2 x
Mihaly Bencze
PP29071. If ak > 0 (k = 1, 2, ..., n) , thenP
1≤i<j≤n
(2a2i−3aiaj+2a2j)(ai+aj)
a4i+a4j≤ (n− 1)
nPk=1
1ak.
Mihaly Bencze
PP29072. If ak > 0 (k = 1, 2, ..., n) , then determine all λ > 0 for whichnP
k=1
ak ≥ λ n
snQ
k=1
ak + (n− λ)
s1n
nPk=1
a2k.
Mihaly Bencze
PP29073. Compute∞R0
dx(1+x4)(1+xe)(1+xπ)
.
Mihaly Bencze
PP29074. If A =
�λ+ 1 λ2 + λ+ 1−1 −λ
�where λ ∈ C then
A− I2 −A2018 = O2.
Mihaly Bencze
PP29075. Compute limn→∞
1n3
P1≤i<j<k≤n
�(n3+i3)(n3+j3)(n3+k3)
n6+i3+j3+k3
� 13
.
Mihaly Bencze
PP29076. Compute limn→∞
Innn when In =
nR0
�xn + xn+1
�arctgxdx.
Mihaly Bencze
Proposed Problems 945
PP29077. Determine all a, b ∈ R for whichP
cyclic
xa1+xb
2x2+x3+...+xn
≥ 2nn+1 .
Mihaly Bencze
PP29078. Prove thatnP
k=1
hqn− 1
k(k+1) +qn+ 1
k(k+1) +√ni= n
�√9n− 2
�when [·] denote the
integer part.
Mihaly Bencze
PP29079. Solve the equationnP
k=1
hx
(3k−1)(3k+1)(2k+2)
i= n when [·] denote
the integer part.
Mihaly Bencze
PP29080. Solve the equationnP
k=1
hx2k−1+x2k+1
(2k−1)(2k+1)
i= n+ 1 when [·] denote the
integer part.
Mihaly Bencze
PP29081. If x, y, z > −1 thenP x4+4x3+16x2+24x+16
x3+4x2+6x+4≥ 12.
Mihaly Bencze
PP29082. Let be M(a, b, c) a common point of the plane x+ y + z = λ1,and the sphere x2 + y2 + z2 = λ2
2 not lying on the cartesian axis. Determineall λ1,λ2 ∈ R for which
P aba+b does not depend of the position of point M.
Jose Luis Diaz-Barrero and Mihaly Bencze
PP29083. Prove thatp−1Pk=0
(−1)k cos�kπ+1
p
�∈ R\Q for all p prime numbers.
Mihaly Bencze
PP29084. In all triangle ABC holds
1).Q
abc ≤�
4s3
9(s2+r2+4Rr)
�9(s2+r2+4Rr)
2).Q
(s− a)(s−b)(s−c) ≤�
s3
9r(4R+r)
�9r(4R+r)
Mihaly Bencze
946 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP29085. Solve in Z the equation2 (3x+ 4y + z)3 + 8 + 4yz + 3x+ z = 4y2 + z2.
Mihaly Bencze and Stanciu Neculai
PP29086. Compute limn→∞
(n!)2nQ
k=1
��1 + 1
k
� 1k+1 − 1
�.
Mihaly Bencze
PP29087. Prove thatnP
k=1
arctg 2k+1(k3+3k+1)2
arctg 2k2+6k+3k4+6k3+11k2+6k−1
=
(arctg3)2 − (arctg (n+ 1) (n+ 3))2 .
Mihaly Bencze
PP29088. Solve in R the following system:
p18 + 3x− y2 +
√9− z2 +
p9− 6x+ y2 +
√9z − 3x2 ≤ 12p
18 + 3y − z2 +√9− x2 +
p9− 6y + z2 +
p9x− 3y2 ≤ 12√
18 + 3z − x2 +p9− y2 +
√9− 6z + x2 +
p9y − 3z2 ≤ 12
.
Mihaly Bencze
PP29089. In all triangle ABC holdsP tg2 A
2tg2 B
2
tgA2tgB
2+3(tg2 B
2tg2 C
2+tg2 C
2tg2 A
2 )≥ 1
3 .
Mihaly Bencze
PP29090. In all triangle ABC holdsP r2ar
2c
ra(r2br2c+3r(r2b+r2c)ra)≥ 1
3r .
Mihaly Bencze
PP29091. Prove thatrQ
p=1
nPk=1
(kr − p+ 1)! can be perfect r power.
Mihaly Bencze
PP29092. Prove that
�nP
k=1
((3k − 2)!)2��
nPk=1
((3k − 1)!)2��
nPk=1
(3k!)2�
can be a perfect cube.
Mihaly Bencze
Proposed Problems 947
PP29093. Determine all A ∈ M2 (C) for which
An +An−1 =
�2 (2n− 1) 20170 2
�.
Mihaly Bencze
PP29094. If ak > 0 (k = 1, 2, ..., n) andnP
k=1
ak ≥ n then
Pcyclic
a21a1+a22+a23+...+a2n
≥ n.
Mihaly Bencze
PP29095. If A =
2+√3
2
√32
√32√
32
2+√3
2
√32√
32
√32
2+√3
2
then
An = I3 +(3
√3+2)
n−2n
3·2n
1 1 11 1 11 1 1
.
Mihaly Bencze
PP29096. If x ∈�0, π2
�and (cosx)a = sinx, (sinx)b = cosx then compute
a2 + b2.
Mihaly Bencze
PP29097. In all triangle ABC holdsP wa
ma≥ 4√
6(s2−r2−4Rr)
P bcb+c cosA.
Mihaly Bencze
PP29098. Prove that (n+ 1) (n+ 2) ... (kn) is divisible by kn and is notdivisible by kn+1, when n, k ∈ N∗.
Mihaly Bencze
PP29099. If xk > 0 (k = 1, 2, ..., n) andnP
k=1
xk = n, thenP
cyclic
x1√x2 ≤ n.
Mihaly Bencze
948 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP29100. Solve in R the equation
�nP
k=1
xk�+ 1
�
n�
k=1xk
� =nP
k=1
{x}k + 1n�
k=1
{x}x
when [·] and {·} denote the integer, respective the fractional part.
Mihaly Bencze
PP29101. Determine all a ∈ C for whichnP
k=0
�a2n+ aak + 1
�a
= 1a2
�a3n+ a2
a2n+ a
�+ 1
a2
�a2n+ aan+ 1
�+ 1
a
�a2n+ aan+ 1
�a2
.
Mihaly Bencze
PP29102. Determine all a ∈ C for which�anPk=0
(−1)k�ank
��a2n−kan
�ak�a
=�ann
� 12a2.
Mihaly Bencze
PP29103. Solve in Z the equation xa1a2...an − yb1b2...bm = a1a2...anb1b2...bm,where ai, bj ∈ {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} (i = 1, 2, ..., n; j = 1, 2, ...,m) .
Mihaly Bencze
PP29104. Solve the following system
{x1}+�x22
= x3+[x4]
2017
{x2}+�x23
= x4+[x5]
2017−−−−−−−−−−{xn}+
�x21
= x2+[x3]
2017
, when [·]
and {·} denote the integer part, respective the fractional part.
Mihaly Bencze
PP29105. Prove that�2n2r
� rPk=0
(−1)k�nk
��n
2r−k
�=
�2nn
� rPk=0
(−1)k�2rk
��2n−2rn−k
�.
Mihaly Bencze
PP29106. Determine all a ∈ C for whichnP
k=0
�a2nak
�a= 1
a2
�a3na2n
�+ 1
a2
�a2nan
�+ 1
a
�a2nan
�a.
Mihaly Bencze
Proposed Problems 949
PP29107. Prove that
�nP
k=0
�2n2k
�2��
n−1Pk=0
�2n
2k+1
�2�
= 14
�4n2n
�2 − 14
�2nn
�2.
Mihaly Bencze
PP29108. Prove that
�rP
k=0
�n2k
��n
2r−2k
���r−1Pk=0
�n
2k+1
��n
2r−2k−1
��= 1
4
�2n2r
�2 − 14
�nr
�2when r ≥ 1.
Mihaly Bencze
PP29109. Prove that4nPk=0
(−1)k�4nk
�3=
�6n2n
� 4nPk=0
(−1)k�4nk
�2.
Mihaly Bencze
PP29110. Determine all a ∈ C for whichanPk=0
(−1)k�ank
�a+1= (−1)n
�ann
��(a+1)n
n
�.
Mihaly Bencze
PP29111. Compute
�nP
k=0
(−1)k�nk
�3�2
−2nPk=0
(−1)k�2nk
�3.
Mihaly Bencze
PP29112. Let A,B ∈ Mn (C) such that AB = BA and A2013 = In,B2017 = In. Compute rank (A+B) .
Mihaly Bencze
PP29113. Prove that n2(n+1) <
nPk=1
√2k(2k+1)−
√(2k−1)(2k+2)
k < 1+ 12 + ...+ 1
n .
Mihaly Bencze
PP29114. If A,B ∈ M2 (Q) such that AB = BA, detA = −λ,
det�A+
√λB
�= 0, then compute det
�A2 +B2 −AB
�, where λ > 0.
Mihaly Bencze
950 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP29115. Determine A ∈ M3 (R) for which Tr (A) = 9 and
A4 =
�3897 4968621 792
�.
Mihaly Bencze
PP29116. Solve in R the following system
x1 − x22 + x3 = x2 − x23 + x4 = ... = x2n − x21 + x2 = 1.
Mihaly Bencze
PP29117. If x1 = 1 and (1 + xn)xn+1 = xn + 2 for all n ≥ 1 then�xn−1 −
√2� �
xn+1 −√2� �
xn +√2�2
=
=�xn−1 +
√2� �
xn+1 +√2� �
xn −√2�2
for all n ≥ 2.
Mihaly Bencze
PP29118. Compute limn→∞
sin�π2
�√4n2 + 2n+ 3 +
√4n2 + 10n+ 9
��·
· cos�π2
�√4n2 + 2n+ 3−
√4n2 + 10n+ 9
��
Mihaly Bencze
PP29119. Solve the equation A4 − 4A3 + 5A2 =
�40 8020 40
�and compute
An, where n ∈ N .
Mihaly Bencze
PP29120. Determine all A ∈ M2 (C) for which A2 = 4A− 3I2 and computeAn, where n ∈ N .
Mihaly Bencze
PP29121. If√nxn + n+ 1−
√nxn + 1 ≤
√n2 ≤ √
nxn + n−√nxn for all
n ≥ 1 then compute∞Pn=1
�916 − xn
�2.
Mihaly Bencze
Proposed Problems 951
PP29122. Solve in R the following system:
xlg(5x2)1 = 2 · 51−log2 x3
xlg(5x3)2 = 2 · 51−log2 x4
−−−−−−−−−−xlg(5x1)n = 2 · 51−log2 x2
.
Mihaly Bencze
PP29123. Prove thatnP
k=1
(−1)k−1 (nk)k2
≥ 1 + 1√2+ ...+ 1
n√n.
Mihaly Bencze
PP29124. Prove that
1).nP
k=0
��nk
�Lk
� 1k+1 ≥ n+1
√L2n
2).nP
k=0
��nk
�2kLk
� 1k+1 ≥ n+1
√L3n when Ln denote the kth Lucas number.
Mihaly Bencze
PP29125. Prove that
1).nP
k=1
(Lk)1k ≥ n
√Ln+3 − 3
2).nP
k=1
�L2k
� 1k ≥ n
√LnLn+1 − 2 where Lk denote the kth Lucas number.
Mihaly Bencze
PP29126. Prove that
1).nP
k=0
��nk
�Fk
� 1k+1 ≥ n+1
√F2n
2).nP
k=0
��nk
�2kFk
� 1k+1 ≥ n+1
√F3n when Fk denote the kth Fibonacci number.
Mihaly Bencze
PP29127. Prove that:
1).nP
k=1
(F2k+1)1k ≥ n
√F2n+2
2).nP
k=1
(F2k)1k ≥ n
√F2n+1 − 1 when Fk denote the kth Fibonacci number.
Mihaly Bencze
952 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP29128. Prove that�n0
�+q�
n1
�+ 3
q�n2
�+ ...+ n+1
q�nn
�≥ 2.
Mihaly Bencze
PP29129. Prove that
1). F1 +√F2 +
3√F3 + ...+ n
√Fn ≥ n
√Fn+2 − 1
2). F 21 +
pF 22 + 3
pF 23 + ...+ n
pF 2n ≥ n
√FnFn+1 when Fk denote the kth
Fibonacci number.
Mihaly Bencze
PP29130. Determine the function f : R → R for which
nPk=1
xk ≤nP
k=1
5xkf (xk) ≤�
nPk=1
xk
�5
n�
k=1xk
for all xk ∈ R (k = 1, 2, ..., n) .
Mihaly Bencze
PP29131. If xk ∈�18 , 1
�(k = 1, 2, ..., n) then
Pcyclic
logx1
�x24 − 1
64
�≥ 2n.
Mihaly Bencze
PP29132. Solve in C the following system:
x1 (x2x3...xn)k = a1
a2
x2 (x1x3...xn)k = a2
a3−−−−−−−−−xn (x1x2...xn−1)
k = ana1
where ap ∈ R (p = 1, 2, ..., n) and k ∈ N.
Mihaly Bencze
PP29133. If an =�1 + 1
n
�nthen solve in R the following system:
x1√x2x3...xn = a1
x2√x1x3...xn = a2
−−−−−−−−xn
√x1x2...xn−1 = an
.
Mihaly Bencze
Proposed Problems 953
PP29134. Solve in C the following system:
3x+ y + z − 28xy = 7�x2 + 4y2
�
x− y + 2z + 4xz = 2�x2 + z2
�
4x− 4y + z + 12xy = 4x2 + 9y2.
Mihaly Bencze
PP29135. In all triangle ABC holds −6− 2√3 ≤P sinA−3
cosA+2 ≤ −6 + 2√3.
Mihaly Bencze
PP29136. Solve in R the following system:�√x1 − 1 +
√2− x2
� �3√x3 − 1 + 3
√2− x4
�= ...
=�√
xn − 1 +√2− x1
� �3√x2 − 1 + 3
√2− x3
�= 1.
Mihaly Bencze
PP29137. Solve in R the following system: 3|x1+1| − 2 |3x2 − 1|− 3x3 =
= 3|x2+1| − 2 |3x3 − 1|− 3x4 = ... = 3|xn+1| − 2 |3x1 − 1|− 3x2 = 2.
Mihaly Bencze
PP29138. Solve in C the following system:
|z1 − |z2 + 1|| = |z3 + |z4 − 1|||z2 − |z3 + 1|| = |z4 + |z5 − 1||−−−−−−−−−−−−−|zn − |z1 + 1|| = |z2 + |z3 − 1||
.
Mihaly Bencze
PP29139. If x ∈�0, π2
�then�
2sinx + 3− cosx + 4− sinx + 6cosx� �
2cosx + 3− sinx + 4− cosx + 6sinx�≥ 16.
Mihaly Bencze
PP29140. Solve in R the following system:
�x21 + 4
�52x
22−10x3+7 = x4�
x22 + 4�52x
23−10x4+7 = x5
−−−−−−−−−−−�x2n + 4
�52x
21−10x2+7 = x3
.
Mihaly Bencze
954 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP29141. Solve in R the following system:
2x1 + 3x2 + 5x3 + 7x4 = 15x5 + 22x2 + 3x3 + 5x4 + 7x5 = 15x6 + 2−−−−−−−−−−−−−−−2xn + 3x1 + 5x2 + 7x3 = 15x4 + 2
.
Mihaly Bencze
PP29142. Solve in C the following system:
x1x2x3 = x34 + 3x5 − x6 − x7 − x8x2x3x4 = x35 + 3x6 − x7 − x8 − x9−−−−−−−−−−−−−−−xnx1x2 = x33 + 3x4 − x5 − x6 − x7
.
Mihaly Bencze
PP29143. If εn = 1 (ε 6= 1) then
(z − 1)2 + (z − ε)2 +�z − ε2
�2+ ...+
�z − εn−1
�2= nz2 for all z ∈ C.
Mihaly Bencze
PP29144. Solve in R the following system:
1 + log5 x1 = log3�4 +
√5x2
�
1 + log5 x2 = log3�4 +
√5x3
�
−−−−−−−−−−−−−1 + log5 xn = log3
�4 +
√5x1
�.
Mihaly Bencze
PP29145. Solve in R the following system:
√1 + 2x1 +
3√1 + 3x2 +
4√1 + 4x3 +
5√1 + 5x4 = 4 + 4x5√
1 + 2x2 +3√1 + 3x3 +
4√1 + 4x4 +
5√1 + 5x5 = 4 + 4x6
−−−−−−−−−−−−−−−−−−−−−−−−−√1 + 2xn + 3
√1 + 3x1 +
4√1 + 4x2 +
5√1 + 5x3 = 4 + 4x4
.
Mihaly Bencze
PP29146. In all triangle ABC holdsP (a+b)2
3√a2
b2�
3√a2+3
3√c2� ≥ 3.
Mihaly Bencze
Proposed Problems 955
PP29147. Solve in R the following system:
4√sinx1 + 2
√cosx2 = log2
81+x2
3
4√sinx2 + 2
√cosx3 = log2
81+x2
4
−−−−−−−−−−−−−4√sinxn + 2
√cosx1 = log2
81+x2
2
.
Mihaly Bencze
PP29148. Determine the functions f : R → R where f (x) = ax2 + bx+ c
such that
f ◦ f ◦ ... ◦ f| {z }
n−time
(x) +
f ◦ f ◦ ... ◦ f| {z }
n−time
(−x) = 0 for all x ∈ R.
Mihaly Bencze
PP29149. Solve in R the following system:
x21 +3px62 + 7x33 =
px44 + 8x5
x22 +3px63 + 7x34 =
px45 + 8x6
−−−−−−−−−−−−−−x2n + 3
px61 + 7x32 =
px43 + 8x4
.
Mihaly Bencze
PP29150. Solve in R the following system:
2sin 3x1 = sin3 x2 + 8sinx3
2sin 3x2 = sin3 x3 + 8sinx4
−−−−−−−−−−2sin 3xn = sin3 x1 + 8sinx2
.
Mihaly Bencze
PP29151. We consider the sequence (an)n≥1 such that a1 = 0, a2 = 1 andall an, an+1, an+2 for n odd are in arithmetical progression and for n even are
in geometrical progression. Prove thatnP
k=2
k2
a2k−1a2k= n−1
n .
Mihaly Bencze
PP29152. If ak =h√
k2 − k + 1 +√k2 + k + 1
iwhen [·] denote the integer
part, then 9 (n+ 1)2�
nPk=1
akak+1
�2
= 4
�nP
k=1
a3k
��n+1Pk=1
a3k
�.
Mihaly Bencze
956 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP29153. In all triangle ABC holdsPq
9�tg2A2 + tg2B2
�+ ctg2C2 ≥ 9.
Mihaly Bencze
PP29154. Solve in R the following system:
x1 (x1 + x2 + ...+ xn)k = 13
x2 (x1 + x2 + ...+ xn)k = 23
−−−−−−−−−−−−xn (x1 + x2 + ...+ xn)
k = n3
when k ∈ N.
Mihaly Bencze
PP29155. If a, b, c ≥ 0, a2 + b2 + c2 = 1, thenP a2
1+2014bc ≥ 32017 .
Mihaly Bencze
PP29156. Computeb−1Pk=1
nk2ab
owhere a, b ∈ N∗ and {·} denote the
fractional part.
Mihaly Bencze
PP29157. Prove that the number n4 + 49n2 + 2027 can be the sum of twoprime number, for all n ∈ N.
Mihaly Bencze
PP29158. If {a1, a2, ..., an} =�13, 23, ..., n3
then solve the following
system:
x1
�nP
k=1
xk
�= a1
x2
�nP
k=1
xk
�= a2
−−−−−−−−xn
�nP
k=1
xk
�= an
.
Mihaly Bencze
PP29159. Let be a0 = x and 5an+1 = a2n − 10an + 50 for all n ≥ 0. Provethat exist infinitely many x ∈ R\Q for which an ∈ N.
Mihaly Bencze
Proposed Problems 957
PP29160. Let AkBkCk triangles for which BkAkCk∡ ≥ 90 (k = 1, 2, ..., n)and denote hk the altitude from Ak to side BkCk (k = 1, 2, ..., n) . Prove thatnQ
k=1
1hk
≥nQ
k=1
1AkBk
+nQ
k=1
1AkCk
.
Mihaly Bencze
PP29161. If ma (x, y) =x+y2 , mg (x, y) =
√xy, mh = 2xy
x+y ,
mp (x, y) =q
x2+y2
2 then solve in R the following system:
mh (x1, x2) +mp (x2, x3) = mg (x3, x4) +ma (x4, x5)mh (x2, x3) +mp (x3, x4) = mg (x4, x5) +ma (x5, x6)−−−−−−−−−−−−−−−−−−−−−−−mh (xn, x1) +mp (x1, x2) = mg (x2, x3) +ma (x3, x4)
.
Mihaly Bencze
PP29162. Determine all x ∈ R\Q for which x (x+ 1) and x2 (x+ 2017) areinteger numbers.
Mihaly Bencze
PP29163. If ma (x, y) =x+y2 ,mg (x, y) =
√xy,mh (x, y) =
2xyx+y ,
mp (x, y) =q
x2+y2
2 then solve in R the following system:
mh (x1, x2) +ma (x2, x3) = mg (x3, x4) +mp (x4, x5)mh (x2, x3) +ma (x3, x4) = mg (x4, x5) +mp (x5, x6)−−−−−−−−−−−−−−−−−−−−−−−mh (xn, x1) +ma (x1, x2) = mg (x2, x3) +mp (x3, x4)
.
Mihaly Bencze
PP29164. If x ∈ R then2 +
h�1 +
h1
sin2 x
i�cos2 x
i+��1 +
�1
cos2 x
��sin2 x
�=h
1sin2 x
i+�
1cos2 x
�when
[·] denote the integer part.
Mihaly Bencze
PP29165. If a is odd and natural number not a perfect cube, then for alln,m ∈ N∗ holds {m (a+ 3
√a)} 6= {n (a− 3
√a)} where {·} denote the
fractional part.
Mihaly Bencze
958 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP29166. Let (an)n≥1 be an arithmetical progression formed by naturalnumbers, and M = {k ∈ N∗|aak = ak+1 + ak + ak−1} . Prove that if2019 ∈ M and M 6= ∅ then M = N∗.
Mihaly Bencze
PP29167. In all triangle ABC holds s2+(4R+r)2
4sR ≥√3 + 3s
2(4R+r) .
Mihaly Bencze
PP29168. If
yn =(Lk)
n+1n +(Lk+1)
n+1n
Lk+2+
(Lk+1)n+1n +(Lk+2)
n+1n
Lk+3+
(Lk+2)n+1n +(Lk)
n+1n
Lk+Lk+2, where Lk
denote the kth Lucas number, then yn−1 + yn+1 ≤ 2yn for all n≥ 2 and k ≥ 1.
Mihaly Bencze
PP29169. Determine all ak > 0 (k = 1, 2, ..., n) if a1 = 1 anda5113
+a5223
+ ...+ a5nn3 = 2n+1
6 anan+1.
Mihaly Bencze
PP29170. If xn =Fn+1k
+Fn+1k+1
Fnk+Fn
k+1+
Fn+1k+1 +Fn+1
k+2
Fnk+1+Fn
k+2+
Fn+1k+2 +Fn+1
k
Fnk+2+Fn
kwhen Fk denote the
kth Fibonacci number, then xn−1 + xn+1 ≤ 2xn for all n ≥ 2 and k ≥ 1.
Mihaly Bencze
PP29171. Prove that (an)n≥1 is an arithmetical progression if and only if2 (a2 + a3 + ...+ an−1) = (n− 2) (a1 + an) for all n ≥ 3.
Mihaly Bencze
PP29172. Solve in R the equationnP
k=1
{kx} = 1 when {·} denote the
fractional part.
Mihaly Bencze
PP29173. Solve in R the equationnP
k=1
�k2x
= 1 when {·} denote the
fractional part.
Mihaly Bencze
Proposed Problems 959
PP29174. If xk > 0 (k = 1, 2, ..., n) such thatnP
k=1
1xk
≤ n thennP
k=1
x2k ≥ n.
Mihaly Bencze
PP29175. Prove thatnP
k=3
p25 + (k − 3) (k − 1) (k + 2) (k + 4) = 2n3+6n2−39n+36
6 .
Mihaly Bencze
PP29176. Prove thathpn (n+ 1) +
pn (n+ 2) +
p(n+ 1) (n+ 2)
i= 3n+ 2 for all n ∈ N∗,
where [·] denote the integer part.
Mihaly Bencze
PP29177. If ak > 0 (k = 1, 2, ..., n) , a1 = 1 and1
a1+a2+ 1
a2+a3+ ...+ 1
an−1+an= an − 1 for all n ≥ 1 then
2�
1an+1
− 1�< a1 + a2 + ...+ an < 2
an.
Mihaly Bencze
PP29178.Prove thatnP
k=1
h(k+1)2
k+2
i h(k+2)2
k+3
i= n(n+1)(n+2)
3 where [·] denote the
integer part.
Mihaly Bencze
PP29179. If r, n,m, p ∈ N ∗ then
r+1Pk=1
hkn+1
kn+kn−1+...+k+1
i hkm+1
km+km−1+...+k+1
i hkp+1
kp+kp−1+...+k+1
i= r2(r+1)2
4 , where
[·] denote the integer part.
Mihaly Bencze
PP29180. If x ∈ R then1+27tgx
3(1+3tgx+9ctg4x)+ 1+27ctgx
3(1+3ctgx+9tg4x)+
81(tg3x+tg3y)1+81(tgx+ctgx) ≥ 2.
Mihaly Bencze
960 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP29181. Determine all r, s ∈ R\Q for which�rs
�n+�sr
�n ∈ N for alln ∈ N.
Mihaly Bencze
PP29182. Prove that
π2R0
2n√cosxdx2n√sinx+ 2n√cosx
= π4 when n ∈ N∗.
Mihaly Bencze
PP29183. Determine all n ∈ N for whichnP
k=1
k5nk is divisible by 5.
Mihaly Bencze
PP29184. Solve in R the equationp3 (2x+ 3) (3y + 5) +
p6 (x+ 2) (y + 1) +
p2 (x+ 1) (3y + 1) =
=p
5 (2x+ 3) (9y + 8).
Mihaly Bencze
PP29185. If x, y > 0 and x2 + y2 ≤ 1 then1√
x2+2√2y+3
+ 1√y2+2
√2x+3
+ 1√9+xy
≥ 1.
Mihaly Bencze
PP29186. Let f : R → R be a continuous convex function andλ ∈ [0, 1] , a, b > 0. Prove that
λa
aR0
f (x) dx+ 1−λb
bR0
f (x) dx ≥ 1λa+(1−λ)b
λa+(1−λ)bR0
f (x) dx.
Mihaly Bencze
PP29187. Let f : R → R be a continuous function and
λk > 0, ak > 0 (k = 1, 2, ..., n) such thatnP
k=1
λk = 1. Prove that
nPk=1
λk
ak
akR0
f (x) dx ≥ 1n�
k=1
λkak
n�
k=1λkakR0
f (x) dx.
Mihaly Bencze
Proposed Problems 961
PP29188. Determine all k ∈ N∗ for which (k − 1)! is divisible by k+1.
Mihaly Bencze
PP29189. If P0 = 0, P1 = 1 and Pn = 2Pn−1 + Pn−2 for all n ≥ 2, then
compute limn→∞
nPk=1
arctg2(Pk+Pk+1)Pk+1
PkP2k+1Pk+2−1
· arctg�
2P 2k+1
1+PkP2k+1Pk+2
�.
Mihaly Bencze and Jose Luis Dıaz-Barrero
PP29190. Determine all p ∈ N∗ such that limn→∞
nPk=1
12n+pk−1 = ln
√2.
Mihaly Bencze
PP29191. In all triangle ABC holds1).
P aa+2b < 4
3 − 23Rr6s2
2).P −a+b+c
a−b+3c < 43 − 23r2
3s
3).P ra
ra+2rb< 4
3 − 23s2r3(4R+r)3
Mihaly Bencze
PP29192. If a1 = 1 and an+1 (1 + nan) = an for all n ≥ 1, and
λ = limn→∞
1n4
�1a1
+ 2a2
+ ...+ nan
�, then compute n
�λ− 1
n4
nPk=1
kak
�.
Mihaly Bencze
PP29193. Determine all A,B,C ∈ M3 (C) for whichdet
�A3 +B3 + C3
�= 1, AB = C,BC = A,CA = B and A2 +B2 + C2 = I3.
Mihaly Bencze
PP29194. Determine sinAk the subunitary root of the equation2 (ak + 2)x2 − 2 (ak + 1) (ak + 1)x− ak = 0, where k ∈ {1, 2, 3}. Determineall a1, a2, a3 ∈ R for which A1 +A2 +A3 = π.
Mihaly Bencze
PP29195. Compute
1).1R0
arctgxx+1 dx 2).
1R0
arctgx·arctg(x+1)(x+1)(x+2) dx
Mihaly Bencze
962 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP29196. Solve in Z the equation x3
3y+5z + y3
3z+5x + z3
3x+5y = 32 .
Mihaly Bencze
PP29197. If xn = 1k
k + 2k
k2+ ...+ nk
kn when k ∈ N, k ≥ 2, then compute [xn],where [·] denote the integer part.
Mihaly Bencze and Nicolae Papacu
PP29198. If xn > 0, n ∈ N∗ andnP
k=1
xk+1
k2xk< n
n+1 , then the sequence (xn)n≥1
is convergent.
Mihaly Bencze
PP29199. Solve in R the following system:
loga x+ loga+1 (y + 1) = 2 loga+2 (z + 2)loga y + loga+1 (z + 1) = 2 loga+2 (x+ 2)loga z + loga+1 (x+ 1) = 2 loga+2 (y + 2)
.
Mihaly Bencze
PP29200. Let ABC be a triangle, then determine all x, y > 0 for which16P �
cos A2
�x+y �sin B
2
�x �sin C
2
�y ≥ 3sR .
Mihaly Bencze
PP29201. Determine all a, b, c ∈ R for which the equationsx2 + (a+ c)x+ b2 − c = 0, x2 + (b+ a)x+ c2 − a = 0,x2 + (c+ b)x+ a2 − b = 0 have real roots.
Mihaly Bencze
PP29202. Prove thatnP
k=1
tg2 π3(2k+1) ≤
n(16n2+48n+17)3 .
Mihaly Bencze
PP29203. Prove thatnP
k=1
ln(1+√k−1)
k2(k+1)< n
2(n+1) .
Mihaly Bencze
Proposed Problems 963
PP29204. Determine all a ∈ N, a ≥ 2 for which an − 1 is divisible bya(n+ 1) for an infinity of n ∈ N.
Mihaly Bencze
PP29205. Determine all a, b, c ∈ Z for which a+b+c−abc1+ab+bc+ca ∈ Z.
Mihaly Bencze
PP29206. Prove thatnP
k=1
sin πk(k+1)n ≥ n
n+1 sinπn
Mihaly Bencze
PP29207. If zk ∈ C∗ (k = 1, 2, ..., n) such that|z1| = |z2| = ... = |zn| = |z1 + z2 + ...+ zn| then compute
card
��z1z2
�t+�z2z3
�t+ ...+
�znz1
�t|t ∈ N
�.
Mihaly Bencze
PP29208. If k ∈ N ∗ then solve in R the equation
[kx] +�
1kx
= {(k + 1)x}+
h1
(k+1)x
iwhere [·] , respective {·} denote the
integer respective the fractional part.
Mihaly Bencze
PP29209. If ak ∈ Q (k = 1, 2, ..., 2n) such thatan = 1
n (a1 + a3 + ...+ a2n−1) and an+1 =1n (a2 + a4 + ...+ a2n) , then
compute2nPk=1
1a2k
.
Mihaly Bencze
PP29210. Determine all prime p and all k ∈ N for which (p+ 1)k+1 − pk+1
is a perfect k power.
Mihaly Bencze
PP29211. Solve in Z the equation (x− y) (1 + xy) =�1− x2
� �1− y2
�.
Mihaly Bencze
964 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP29212. Let ABC be a triangle such that x cosA+ y cosB = 1 andy sinA+ x sinC = (x+ y) sinA. Determine all x, y > 0 for which ABC isequilateral.
Mihaly Bencze
PP29213. In all triangle ABC holds a2x+ b2y + c2z ≤ 4p
x2 + y2 + z2srfor all x, y, z > 0.
Mihaly Bencze
PP29214. In all triangle ABC holdsP
a2 ≥ sr
�4√
�
xy�
x
�for all x, y, z > 0.
Mihaly Bencze
PP29215. In all triangle ABC holds�a2xy + b2yz + c2zx
�2 �a2x+ b2y + c2z
�2 ≥≥ 256xyz (x+ y + z) (xy + yz + zx) (sr)4 for all x,y,z¿0.
Mihaly Bencze
PP29216. In all triangle ABC holds
a2eb+c + b2ec+a + c2ea+b ≥ 4√3sre
2(a+b+c)3 .
Mihaly Bencze
PP29217. In all triangle ABC holds�P
a2b� �P
a2c�≥ 16 (
Pab)2 (sr)2 .
Mihaly Bencze
PP29218. In all triangle ABC holds�P
a2bλ� �P
a2cλ�≥ 16
�P(ab)λ
�2(sr)2 for all λ ≥ 0.
Mihaly Bencze
PP29219. In all triangle ABC holds1).
Pa ≥ 4s
pr2R
2).P
a2 ≥ 4√3sr
3).P
a3 ≥ 4√s2 + r2 + 4Rrsr
4).P
a2 (s− b) ≥ 4pr (4R+ r)sr
5).P
a2rb ≥ 4s2r
6).P �
a sin A2
�2 ≥ srR
√s2 + r2 − 8Rr
Proposed Problems 965
7).P �
a cos A2
�2 ≥ srR
qs2 + (4R+ r)2
Mihaly Bencze
PP29220. In all triangle ABC holdsP
a2 sinA ≥ 2srR
√s2 + r2 + 4Rr.
Mihaly Bencze
PP29221. In all acute triangle ABC holdsP
a2 cosA ≥ 2srR
√s2 + r2 − 4R2.
Mihaly Bencze
PP29222. In all triangle ABC holds a2ea + b2eb + c2ec ≥ 4√3sre
a+b+c3 .
Mihaly Bencze
PP29223. In all triangle ABC holds
r2 + r2a + r2b + r2c ≥ 16R2 − 31−λ�a
2λ + b
2λ + c
2λ
�λfor all λ ∈ [0, 1] .
Mihaly Bencze
PP29224. In all triangle ABC holds1).
Pa2tgA
2 ≥ 4sr
2).P
a2ma ≥ 4s2rq
2rR
Mihaly Bencze
PP29225. In all triangle ABC holds 36r2 ≤ (P
HA) (P
IA) ≤ 6R (2R− r)(A new refinement of Euler’s R ≥ 2r inequality).
Mihaly Bencze
PP29226. In all triangle ABC holdsP
HAλ +P
IAλ ≤ 3
�Rλ +
�2(2R−r)
3
�λ�
for all λ ∈ [0, 1] .
Mihaly Bencze
PP29227. Let ABC be a triangle and denote F the area of the triangleformed by sides ma,mb,mc. Prove that
F ≤ 5(s2−r2−4Rr)2−3(s2+r2+4Rr)
2+48s2Rr
16sr .
Mihaly Bencze
966 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP29228. In all acute triangle ABC holdsP �
cosAsinB sinC
�λ ≤ 3�23
�λfor all
λ ∈ [0, 1] .
Mihaly Bencze
PP29229. Let ABC be a triangle and denote F the area of the triangleformed by sides cos A
2 , cosB2 , cos
C2 . Prove that
F ≤ 116sr
P �b2 + c2 − a2
�cos2 A
2 .
Mihaly Bencze
PP29230. Consider the triangles AkBkCk with sides ak, bk, ck and areasFk = Area [AkBkCk] (k = 1, 2, ..., n) , then
16
�nQ
k=1
Fk
� 2n
+ 2
�nQ
k=1
ak
� 4n
+ 2
�nQ
k=1
bk
� 4n
+ 2
�nQ
k=1
ck
� 4n
≤
≤nQ
k=1
�a2k + b2k + c2k
� 2n for all n ∈ N∗, n ≥ 2 (A generalization of Padoe’s
inequality).
Mihaly Bencze
PP29231. In all triangle ABC holds
1).P
a2u ≤ 31−u�8R2 + 4sr
3√3
�u
2).��
8R2�v
+�
4sr3√3
�v� 1v ≥ 2
1v−1
�Pa2�for all u ∈ [0, 1] and v ≥ 1.
Mihaly Bencze
PP29232. In all triangle ABC holdsP
aλ ≥ 31−λ (2s)λ ≥ 31−λ2
�4√3Area [ABC]
�λ2 for all λ ≥ 1.
Mihaly Bencze
PP29233. In all triangle ABC holds s2λ + 5r2λ ≥ 6�8Rr3
�λfor all λ ≥ 0.
Mihaly Bencze
PP29234. In all triangle ABC holds 2Rλ +�3√3− 4
�rλ ≥
�3√3− 2
�1−λsλ
for all λ ≥ 0.
Mihaly Bencze
Proposed Problems 967
PP29235. In all triangle ABC holdsxaλ
b+c +ybλ
c+a + zcλ
a+b ≥ (xa+yb+zc)λ
(x+y+z)λ−2((y+z)a+(z+x)b+(x+y)c)for all x, y, z > 0 and
λ > 0.
Mihaly Bencze
PP29236. In all triangle ABC holds wλa + wλ
b +mλc ≤ 31−
λ2 sλ for all
λ ∈ [0, 1] .
Mihaly Bencze
PP29237. In all triangle ABC holdsP 1
hλa≥ 3
(3r)λ≥ 3
�3
qR
2s2r2
�λ
for all
λ > 0.
Mihaly Bencze
PP29238. In all triangle ABC holds
3 (3r)λ ≤ 3�
3√s2r
�λ≤P
�a cos A
2
�λ ≤ 31−λ�2�s2 − r2 − 4Rr
��λ2�4R+r2R
�λ2 ≤
≤ 3Rλ�4R+r2R
�λ2 ≤ 3
�3R2
�λfor all λ ∈ [0, 1] .
Mihaly Bencze
PP29239. In all triangle ABC holds
2 +P √
s−a+√s−b√
s−a≤ 2
s2r
q2 (s2 −m2
a)�s2 −m2
b
�(s2 −m2
c).
Mihaly Bencze
PP29240. In all triangle ABC holds
s+ 3Pp
(s− a) (s− b) ≤ 2s
Pq(s2 −m2
a)�s2 −m2
b
�.
Mihaly Bencze
PP29241. In all acute triangle ABC holdsP
(cos (B − C))λ ≤ 31−λ�P ha
ma
�λfor all λ ∈ [0, 1] .
Mihaly Bencze
PP29242. In all triangle ABC holds√2Pp
s (s− a) ≤Pps2 −m2
a.
Mihaly Bencze
968 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP29243. In all triangle ABC holds 18 ≤ s2+r2+4RrRr ≤ 9R
r (A newrefinement of Euler’s R ≥ 2r inequality).
Mihaly Bencze
PP29244. In all triangle ABC holdsP
(ab)2λ ≥ 3�8s2Rr
3
�λ≥ 3
�163 Area [ABC]
�λfor all λ > 0.
Mihaly Bencze
PP29245. In all triangle ABC holds
3�2s2r2
R
�λ3 ≤Phλa ≤Pmλ
a ≤ 31−λ�2s−
�6√3− 9
�r�λ ≤
≤ 31−λ (4R+ r)λ ≤ 31−λ�9R2
�λfor all λ ∈ [0, 1] .
Mihaly Bencze
PP29246. In all triangle ABC holds�2s3
�λ ≤�
(a sin A2 )
λ
�
(sin A2 )
λ ≤ 12 max
naλ + (b+ c)λ , bλ + (c+ a)λ , cλ + (a+ b)λ
ofor
all λ ≥ 0.
Mihaly Bencze
PP29247. In all triangle ABC holds7R2 ≤ (4R+ r)2 + r2 − 2s2 = r2 +
Pr2a ≤ 16R2 − 4s2
3 .
Mihaly Bencze
PP29248. In all acute triangle ABC holdsP
(a cosA)λ ≤ 31−λ�2srR
�λ ≤ 31−λ�
2s2
3√3R
�λ≤ 31−λsλ for all λ ∈ [0, 1] .
Mihaly Bencze
PP29249. In all triangle ABC holds Rλ + rλ ≤ 21−λmax�hλa , h
λb , h
λc
for all
λ ∈ [0, 1] (A generalization of Erdos inequality).
Mihaly Bencze
PP29250. In all triangle ABC holds a2b2c2 ≥ 4s2
27R2a2b2c2 ≥
�4Area[ABC]√
3
�3.
Mihaly Bencze
Proposed Problems 969
PP29251. In all acute triangle ABC holds s2−(2R+r)2
4R2 =Q
cosA ≤ r2
2R2 ≤ 18 .
Mihaly Bencze
PP29252. In all triangle ABC holds1). 512s2Rr ≤ 27
�s2 + r2 + 2Rr
�2
2). 4 (4R+ r)3 r ≤ 27s2R2
3). 16�s2 + r2 + 4Rr
�3 ≤ 27s2�s2 + r2 + 2Rr
�2
4). 512 (2R− r)3 r2 ≤ 27�(2R− r)
�s2 + r2 − 8Rr
�− 2Rr2
�2
5). 512 (4R+ r)3 s2 ≤ 27�(4R+ r)3 + s2 (2R+ r)
�2
Mihaly Bencze
PP29253. In all triangle ABC holdsP 1
(sin A2 )
2n ≥P 1
(sin A2sin B
2 )n ≥ 3
�4Rr
� 2n3 ≥ 3 · 4n for all n ∈ N.
Mihaly Bencze
PP29254. In all triangle ABC holds 18 ≥Q sin A+B
4 ≥ 14
pr2R . (A
refinement of Euler’s R ≥ 2r inequality).
Mihaly Bencze
PP29255. If x, y, z ∈�0, π4
�then
tgxtgytgztg3�x+y+z
3
�≤ tg2
�x+y2
�tg2
�y+z2
�tg2
�z+x2
�.
Mihaly Bencze
PP29256. In all triangle ABC holdsP 1
(sin A2 )
n ≥ 2P 1
(sin A+B4 )
n − 3 · 2n ≥ 3 · 2n for all n ∈ N.
Mihaly Bencze
PP29257. If 2 < a ≤ b thenbRa
8x2−12x+1√x2−2x
dx ≥ 4 (b− a) (b+ a− 1) .
Mihaly Bencze
PP29258. If 2 < a ≤ b thenbRa
8x3−28x2+25x−2(x−2)
√x−2
dx ≥ 2 (b− a) (b+ a− 2) .
Mihaly Bencze
970 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP29259. In all triangle ABC holds1).
Pha cosA ≤ s2−r2−4Rr
2R
2).P
ra cosA ≤ s2−2r(4R+r)2r
3).P
sin2 A2 cosA ≤ 1
2
�s2+r2−8Rr
4Rr
�2+ r
2R − 1
4).P
cos2 A2 cosA ≤ (s2+(4R+r)2)
2−16Rs2(4R+r)
32s2R2
Mihaly Bencze
PP29260. In all triangle ABC holds 2P
a�b2 + c2
�≥ 9abc+
Pa3.
Mihaly Bencze
PP29261. In all triangle ABC holdsP
am2a = s
2
�s2 + 5r2 + 2Rr
�.
Mihaly Bencze
PP29262. In all triangle ABC holdsP a2λ
mbmc≥ 4λ91−λ (
Pmamb)
λ−1 for allλ > 0.
Mihaly Bencze
PP29263. In all triangle ABC holdsP�
a3
ra
�λ≤ 3
�4sR3
�λfor all λ ∈ [0, 1] .
Mihaly Bencze
PP29264. Let ABC be a triangle in which A∡ = 90◦ and x, y > 0. Provethat xyha ≤ (xb+ yc)min {x, y}++1
2
�x2 + y2 −min
�x2, y2
− 2min {x, y}
px2 + y2
�a.
Mihaly Bencze
PP29265. In all acute triangle ABC holdsP
(sinA)λ12 (cosB)
λ22 ≤
≤r
32−λ1−λ2
�2 cos A
2 − 3√3
2
�λ1 �2P
sin A2 − 3
2
�λ2 ≤ 31+λ14
2λ1+λ2
2
for all
λ1,λ2 ∈ (0, 1) .
Mihaly Bencze
PP29266. In all triangle ABC holdsP aλ+1
ra≤ abc
r
�12s
Paλ−1
�for all λ ≥ 0.
Mihaly Bencze
Proposed Problems 971
PP29267. In all triangle ABC holds 4 ≤�
s2+r2−4R2
s2−(2R+r)2
�2− 8R(R+r)
s2−(2R+r)2≤
�Rr
�2
(A new refinment of Euler’s R ≥ 2r inequality).
Mihaly Bencze
PP29268. Solve in Q the equation(11x + 12x + 13x + 14x) (31y + 33y + 35y + 37y + 39y + 41y) = 1320x+y.
Mihaly Bencze
PP29269. Solve in Q the equation(3x + 4x + 5x) (11y + 12y + 13y + 14y) (31z + 33z + 35z + 37z + 39z + 41z) == 7920x+y+z.
Mihaly Bencze
PP29270. If a, b, c > 0 then 2abc+ (P
a)�P
a2�≥ 11
27 (P
a)3 .
Mihaly Bencze
PP29271. Solve in Q the equation(1x + 2x + ...+ 24x) (15y + 16y + ...+ 34y) = 70x+y.
Mihaly Bencze
PP29272. If x ≥ 0 then�524288x16 + 2048
�16 ≥ 256(2048x16+256x15+4480x13+17472x11+22880x9++11440x7 + 2184x5 + 140x3 + 2x− 8)16.
Mihaly Bencze
PP29273. If x ≥ 0 then 9�3x3 + 3x2 + 2x+ 1
�3+9
�9x3 + 6x2 + 3x+ 1
�3 ≥≥
�3x3 + 6x2 + 5x+ 1
�3+�9x3 + 15x2 + 6x+ 1
�3.
Mihaly Bencze
PP29274. If x, y ≥ 0 then 4�3x3y + 9y4
�3+ 4y12 ≥
�10x4 + 9xy3
�3.
Mihaly Bencze
PP29275. If x ≥ 0 then�272x8 + 17
�8 ≥ 17�16x8 + 16x7 + 56x5 + 28x3 + 2x− 1
�8.
Mihaly Bencze
972 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP29276. Prove that maxx,y∈[0,1]
�√x2+1−
√y2+1
�2
(x2+1)(y2+1)≥ π
4 − 12e
1−π4 .
Mihaly Bencze
PP29277. If a ∈ C then�a4 + 2ai
�3+�1 + a3i
�3+�3a2 + 5ai− 5
�3+�4a2 − 4ai+ 6
�3+
+�5a2 − 5ai− 3
�3=
�1 + 2a3i
�3+�a4 − ai
�3+�6a2 − 4ai+ 4
�3.
Mihaly Bencze
PP29278. Prove that∞Pk=0
k!(n+k)! =
1(n−1)(n−1)! for all n ≥ 2.
Mihaly Bencze
PP29279. Prove that
� ∞Pk=0
cos kxk!
�2
= e2 cosx +
� ∞Pk=0
sin kxk!
�� ∞Pk=0
(−1)k sin kxk!
�, for all x ∈ R.
Mihaly Bencze
PP29280. Prove that for all n ∈ N holds 2he2+12e (2n)!
i= [e (2n)!] +
h(2n)!e
i
when [·] denote the integer part.
Mihaly Bencze
PP29281. Prove that [e (2n+ 2)!]−h(2n+2)!
e
i= 4 (n+ 1)
he2−12e (2n+ 1)!
i
for all n ∈ N where [·] denote the integer part.
Mihaly Bencze
PP29282. Prove that1R0
e(sinπ2x)
2n+(cos π
2x)
2n
dx ≤
≤ e2(2n−1)!!
(2n)!! + maxx,y∈[0,1]
qe(sin
π2x)
2n+(cos π
2x)
2n
−q
e(sinπ2y)
2n+(cos π
2y)
2n
!2
.
Mihaly Bencze
Proposed Problems 973
PP29283. Prove that1R0
e(sinπ2x)
2n+1+(cos π
2x)
2n+1
dx ≤ e4(2n)!!
π(2n+1)!!+
+ maxx,y∈[0,1]
qe(sin
π2x)
2n+1+(cos π
2x)
2n+1
−qe(sin
π2y)
2n+1+(cos π
2y)
2n+1
!2
, for all
n ∈ N.
Mihaly Bencze
PP29284. Prove that
1).1R0
ecosπxdx ≤ maxx,y∈[0,1]
�√ecosπx −
√ecosπy
�2+ 1
2).1R0
esinπxdx ≤ maxx,y∈[0,1]
�√esinπx −
√esinπy
�2+ e
2π
Mihaly Bencze
PP29285. Solve in Z the equation x+2yy+3z + y+2z
z+3x + z+2xx+3y = 9
4 .
Mihaly Bencze
PP29286. Prove that1R0
�x+ 1
x
�xdx ≤ 2
√e+ max
x,y∈[0,1]
��x+ 1
x
�x2 −
�y + 1
y
� y2
�2
.
Mihaly Bencze
PP29287. If a ∈ [0, 1] then ea−1a ≤
√ea + max
x,y∈[0,1]
�√eax −
√eay
�2.
Mihaly Bencze
PP29288. Prove that
1). πn (n+ 1) ≤nP
k=1
�k!�ek
�k�2≤ e2n(n+1)
2
2). 2π2n(n+1)(2n+1)3 ≤
nPk=1
�k!�ek
�k�4≤ e4n(n+1)(2n+1)
6
3). 2π3n2 (n+ 1)2 ≤nP
k=1
�k!�ek
�k�6≤ e6n2(n+1)2
4
Mihaly Bencze
974 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP29289. Prove that2(
√n+1−1)e <
nPk=1
�ke
�k · 1k! <
√nπ .
Mihaly Bencze
PP29290. Prove thatnP
k=0
12k+1(n−k)!
=√e
2n+1n!Γ�n+ 1, 12
�.
Mihaly Bencze
PP29291. If x ∈�0, π2
�then
�sin2 x
� �9−sinx3−sinx
� 3sin x
+�cos2 x
� �9−cosx3−cosx
� 3cos x ≥ 82− 18 (sinx+ cosx) .
Mihaly Bencze
PP29292. Prove thatnP
k=1
kq
n2 (n+ 1)2 − k2 (k + 1)2 < πn2(n+1)2
8 .
Mihaly Bencze
PP29293. Solve in Z the equation x+y2
z2+x+ y+z2
x2+y+ z+x2
x+z2= 3.
Mihaly Bencze
PP29294. Solve in Z the equationx(y2+z2)
y+z +y(z2+x2)
z+x +z(x2+y2)
x+y = 3t2.
Mihaly Bencze
PP29295. If xk ≥ 3 (k = 1, 2, ..., n) , then
nQk=1
(3xk−1)xk−2
(xk−1)xk≥
�
3n
n�
k=1xk−1
�
n�
k=1xk−2n
�
1n
n�
k=1xk−1
�
n�
k=1xk
.
Mihaly Bencze
PP29296. If xk ≥ 3 (k = 1, 2, ..., n) , thennP
k=1
�3xk−1xk−1
�xk ≥ 1n
�3
nPk=1
xk − n
�2
.
Mihaly Bencze
Proposed Problems 975
PP29297. If a, b, c ≥ 3 then(3a−1)a−2
(a−1)a· (3b−1)b−2
(b−1)b· (3c−1)c−2
(c−1)c· (a+b+c−1)a+b+c−6
(a+b+c3
−1)a+b+c ≥
≥�
3(a+b)2
−1�a+b−4
( a+b2
−1)a+b ·
�
3(b+c)2
−1�b+c−4
( b+c2
−1)b+c ·
�
3(c+a)2
−1�c+a−4
( c+a2
−1)c+a .
Mihaly Bencze
PP29298. If a, b > 0 and 6a = b�2a2 + 2 + ab
�then 9
8 ≤ 2a2+b2
a(a+2b) ≤65 .
Mihaly Bencze
PP29299. If a, b > 0 and 2 (2a+ b) = a�ab+ 2b2 + a2
�then
2a2�2 + a2
�≥ (1 + ab)
�2 + a2 − b2
�.
Mihaly Bencze
PP29300. Determine all triangles ABC such that2s ≥
�2√3− 1
�R+ 2
�√3 + 1
�r.
Mihaly Bencze
PP29301. If a, b > 0 then�a4 + 2a3b+ b4
� �a4 + 2ab3 + b4
�≥ 4a3b3 (a+ b)2 .
Mihaly Bencze
PP29302. Prove thatnP
k=3
�3k−1k−1
�k≥ 3n(n+1)(2n−1)
2 + n− 29.
Mihaly Bencze
PP29303. In all triangle ABC holds�10R2 −P a2
�Pa3 cosA = 4R2
�Pa2 − 6R2
�Pa cos3A.
Mihaly Bencze
PP29304. If a, b, c ∈ (0, 1] and abc = 1, then 5P
a+ 11+
�
ab ≥ 16.
Mihaly Bencze
PP29305. Determine all a, b ∈ N for which (a+ 1)b + a2 − a+ 1 is aninteger power of 3.
Mihaly Bencze
976 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP29306. In all triangle ABC holdsP
a3 ≥P a3 cos (B − C) ≥ 3abc.
Mihaly Bencze
PP29307. Determine all functions f : R → R for which2f (xy + yz + zx) = f2 (x+ y + z)+ f2 (x)+ f2 (y)+ f2 (z) for all x, y, z ∈ R.
Mihaly Bencze
PP29308. If x, y, z > 0 then
3P
(x+ y)2 ≥�P
x+pP
x2�2
+ 2 (P
x)pP
x2.
Mihaly Bencze
PP29309. If a, b, c > 0 then ln (1 + a) ln (1 + b) ln (1 + c) ln3�1 + 3
√abc
�≤
≤ ln2�1 +
√ab�ln2
�1 +
√bc�ln2 (1 +
√ca) .
Mihaly Bencze
PP29310. Find all function f : N → N such that for all a, b, c ∈ N holds(f (a) + b+ c) (a+ f (b) + c) (a+ b+ f (c)) == (f (a+ b) + c) (f (b+ c) + a) (f (c+ a) + b) .
Mihaly Bencze
PP29311. Prove thatnP
k=1
1k3(k+1)3
≥ 16(√n+1−1)
4
n3(n+1)2.
Mihaly Bencze
PP29312. If a, b > 0 then�a√b+ b
√a+ a
qa+b2
���√a+
√b�√
a+ b+ a√2�+
+
�a√b+ b
√a+ b
qa+b2
���√a+
√b�√
a+ b+ b√2�≤
≤�5a2 + 8ab+ 5b2
�√a+ b.
Mihaly Bencze
PP29313. In all triangle ABC holds arctg π3−ABCπ(π2−
�
AB)≥ π3
4(2π2−�
AB).
Mihaly Bencze
Proposed Problems 977
PP29314. If a0 = 1, b0 = 2, c0 = 3 and an+1 = an + 1bn
+ 1cn,
bn+1 = bn + 1cn
+ 1an, cn+1 = cn + 1
an+ 1
bnfor all n ≥ 0, then compute
max {an, bn, cn} .
Mihaly Bencze
PP29315. In all acute triangle ABC holds�2rR
�2s ≥Q (cos (B − C))a .
Mihaly Bencze
PP29316. Let ABC be a triangle. Determine all x, y > 0 for which2r ≤P 1
wa≤ x
r + yR .
Mihaly Bencze
PP29317. In all acute triangle ABC holds�3√3 +
PctgA
�(P
tgA) ≥ 36.
Mihaly Bencze
PP29318. If a, b, c > 0 thenPq
a+b+2ca+b =
qQ a+b+2ca+b + 4
�
a�
√(a+b)(a+b+2c)
.
Mihaly Bencze
PP29319. If ak ≥ 1 (k = 1, 2, ..., n) and λ ≥ 0, then
(n+ λ+ 1)
�nP
k=1
ak
� Pcyclic
1λa1+a2+a3+...+an
!≤ n2 + λ
P1≤i<j≤n
|ai − aj | .
Mihaly Bencze
PP29320. If x, y, z > 0 thenPq
x+yz ≥
√2 +
q(x+y)(y+z)(z+x)
xyz .
Mihaly Bencze
PP29321. If x, y, z > 0 then
Pqx+yz ≤
q(x+y)(y+z)(z+x)
xyz +√2(x+y+z)
4√xyz(
√x+
√y+
√z).
Mihaly Bencze
978 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP29322. If xk > 0 (k = 1, 2, ..., n) andnP
k=1
xk = n, then
Pcyclic
x31
3�
4(x62+1)
≥ n2 .
Mihaly Bencze
PP29323. If ak > 0 (k = 1, 2, ..., n) and λ ∈ (0, 1] such thatnP
k=1
1ak+λ ≥ 1
then (n− λ− 1)nP
k=1
a2k + λ (n− λ)nP
k=1
ak ≥ 2P
1≤i<j≤naiaj .
Mihaly Bencze
PP29324. If x, y, z > 0 then 2�P
x2�2
+ 12P
xy ≥ 9P
xy2 + 9P
x.
Mihaly Bencze
PP29325. If a, b, c > 0 and abc+P
ab ≤ 4 thenP
a ≥P ab.
Mihaly Bencze
PP29326. Denote Fk and Lk the kth Fibonacci, respective Lucas numbers.Prove that
1).nP
k=1
F 6k+1
(3F 2k−4Fk+3)
2 ≥ 14 (3FnFn+1 − 4Fn+2 + 3n+ 4)
2).nP
k=1
L6k+1
(3L2k−4Lk+3)
2 ≥ 14 (3LnLn+1 − 4Ln+2 + 3n+ 6)
Mihaly Bencze
PP29327. If x ∈�0, π2
�then
1). 1+sin6 x
(3 sin2 x−4 sinx+3)2 + 1+cos6 x
(3 cos2 x−4 cosx+3)2≥ 9− 4 (sinx+ cosx)
2). 1+sin12 x
(3 sin4 x−4 sin2 x+3)2 + 1+cos12 x
(3 cos4 −4 cos2 x+3)2≥ 5− 6 sin2 x cos2 x
Mihaly Bencze
PP29328. If xk > 0 (k = 1, 2, ..., n) thenP 3
√1+x6
1
3x22−4x2+3
≥ n3√4
.
Mihaly Bencze
Proposed Problems 979
PP29329. If 0 < a ≤ b thenbRa
x2dx(3x2−4x+3)3
≥ 112arctg
b3−a3
1+(ab)3.
Mihaly Bencze
PP29330. If 0 < a ≤ b thenbRa
x5dx(3x2−4x+3)3
≥ 124 ln
b6+1a6+1
.
Mihaly Bencze
PP29331. Prove thatnP
k=1
k6+1(3k2−4k+3)2
≥ n(2n2−n+3)8 .
Mihaly Bencze
PP29332. If xk > 0 (k = 1, 2, ..., n) , then
4nnP
k=1
x6k+1
3x2k−4xk+3
≥�3
nPk=1
x2k − 4nP
k=1
xk + 3n
�2
.
Mihaly Bencze
PP29333. If xk > 0 (k = 1, 2, ..., n) , thenP
cyclic
x61+1
(3x21−4x1+3)(3x2
2−4x2+3)2 ≥ n
4 .
Mihaly Bencze
PP29334. If x ∈�0, π2
�, then
1). 32 + 16�sin6 x+ cos6 x
�≥ (9− 4 (sinx+ cosx))3
2). 32 + 16�sin12 x+ cos12 x
�≥
�5− 6 sin2 x cos2 x
�3
Mihaly Bencze
PP29335. In all triangle ABC holdsP 64 sin6 A+9�
4 sin2 A− 8√3sinA+3
�2 ≥ s2−r(4R+r)2R2 − 2s√
3R+ 9
4 .
Mihaly Bencze
PP29336. In all acute triangle ABC holdsP 64 cos6 A+1
(12 cos2 A−8 cosA+3)2≥ 13
4 − 2rR − 3(s2−(2R+r)2)
2R2 .
Mihaly Bencze
980 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP29337. If xk > 0 (k = 1, 2, ..., n) then
nPk=1
�xk +
4(x6k+1)
(3x2k−4xk+3)
2
��x2k + xk + 1
�≥ 3
n3
�nP
k=1
xk
�4
+ 3n
�nP
k=1
xk
�2
+ 3n.
Mihaly Bencze
PP29338. If xk > 0 (k = 1, 2, ..., n) , thenP x2
1
3�
4(x62+1)
≥
�
n�
k=1xk
�2
3n�
k=1x2k−4
n�
k=1xk+3n
.
Mihaly Bencze
PP29339. Solve in (0,+∞) the following system
4(x61+1)
(3x21−4x1+3)
2 = 3x22 − 4x3 + 3
4(x62+1)
(3x22−4x2+3)
2 = 3x23 − 4x4 + 3
−−−−−−−−−−−−−4(x6
n+1)(3x2
n−4x1+3)2= 3x21 − 4x2 + 3
.
Mihaly Bencze
PP29340. If x ∈�0, π2
�then
4 (sinx+ cosx) + 3
q4�1 + sin6 x
�+ 3p4 (1 + cos6 x) ≤ 9.
Mihaly Bencze
PP29341. If x ∈ R then
6 sin2 x cos2 x+ 3
q4�1 + sin12 x
�+ 3p4 (1 + cos12 x) ≤ 7.
Mihaly Bencze
PP29342. Solve in (0,+∞) the following system:
4�x61 + 1
�=
�3x22 − 4x3 + 3
�3
4�x62 + 1
�=
�3x23 − 4x4 + 3
�3−−−−−−−−−−−−−4�x6n + 1
�=
�3x21 − 4x2 + 3
�3
Mihaly Bencze
Proposed Problems 981
PP29343. If xk > 0 (k = 1, 2, ..., n) , then
4n2
�n+
nPk=1
x6k
�≥
�3
nPk=1
x2k − 4nP
k=1
xk + 3n
�3
.
Mihaly Bencze
PP29344. If ak > 0 (k = 1, 2, ..., n) and (a1 + a2) (a2 + a3) ... (an + a1) = 2n
thenQ
cyclic
�1 +
qa1a2(a1+a2)
2
�≤ 2n − 1 +
nQk=1
ak.
Mihaly Bencze
PP29345. In all triangle ABC holds
1).P 1
ra3√
4(729r6+r6a)≥ 1
18r3
2).P 1
ha3√
4(729r6+h6a)
≥ 118r3
Mihaly Bencze
PP29346. In all triangle ABC holdsP (tgA
2tgB
2 )3
�
1+729(tgA2tgB
2 )6≥ 1
9 3√2.
Mihaly Bencze
PP29347. If x ∈�0, π2
�and λ =
√33−sin2 2x−1
2 then�√2 +
qλ�λ+ sin2 x
�sinx
��√2 +
pλ (λ+ cos2 x) cosx
�≤
≤ 2√2(7+λ sin2 x cos2 x)√
2+sinx cosx.
Mihaly Bencze
PP29348. If a, b, c > 0 andP
a+ abc = 4 thenP 1
a+1 ≥ n2−32(n−1) for all
n ∈ N,n ≥ 3.
Mihaly Bencze
PP29349. If a, b > 0 and 2a+ b+ a2b = 4 then a (3b+ 1) ≤ b+ 1.
Mihaly Bencze
982 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP29350. If ak > 0 (k = 1, 2, ..., n) , then
Pcyclic
a1a2+a3+...+an
+
n�
k=1
ak�
cyclic
(a1+a2+...+an−1)≥ n(n−1)n−1+1
(n−1)n.
Mihaly Bencze
PP29351. If a, b > 0 and a+ ab+ b = 12a then 1
3a+4(a3+b3)+ 1
8a3+3b≤ 6
5 .
Mihaly Bencze
PP29352. If a, b > 0 then�4a2+ab+b2
a + aba+b
��4b2+ab+a2
b + aba+b
�≥ 169
16 (a+ b)2 .
Mihaly Bencze
PP29353. If x ∈ R then3�1 + sin4 x
� �1 + cos4 x
�≥ 2
�2 + sin2 x
� �2 + cos2 x
�.
Mihaly Bencze
PP29354. If a, b > 0, then16
�a3 + 3
� �b3 + 3
�≥ (3 + a) (3 + b) (2 + a+ b)2 .
Mihaly Bencze
PP29355. If a, b, c, d > 0 thenP
cyclic
k
qa3+3
b+c+d+1 ≥ 4 for all k ∈ N∗.
Mihaly Bencze
PP29356. If ak > 0 (k = 1, 2, ..., n) , thenP
cyclic
�an−11 +n−1
a2+a3+...+an+1
�λ
≥ n for all
λ ≥ 0.
Mihaly Bencze
PP29357. If x ∈ R then�√32 + cos2 8x− 1
�sin2 x cos2 x
�2 sin4 x+
�√32 + cos2 2x− 1
�cos2 x
�·
·�2 cos4 x+
�√32 + cos2 2x− 1
�sin2 x
��33− 2
√32 + cos2 2x
�≤ 256.
Mihaly Bencze
Proposed Problems 983
PP29358. In all triangle ABC holds:
1).P
(a− b+ c)√b ≤P (a− b+ c)
32
2).P �
cos A2 − cos B
2 + cos C2
�qcos B
2 ≤P�cos A
2 − cos B2 + cos C
2
� 32
3).P
(ma −mb +mc)√mb ≤
P(ma −mb +mc)
32
Mihaly Bencze
PP29359. If a, b > 0 and a (a+ b)2 = 4 then a3b�a2 + b2
�≤ 2.
Mihaly Bencze
PP29360. Let ABCDA1B1C1D1 be a cuboid. Prove that
AB2 ·B1C +BC2 ·B1A+BB21 ·AC ≤
√2�AB3 +BC3 +BB3
1
�.
Mihaly Bencze
PP29361. If ak > 0 (k = 1, 2, ..., n) thenP
cyclic
a21pa22 + a23 ≤
√2
nPk=1
a3k.
Mihaly Bencze
PP29362. If ak > 0 (k = 1, 2, ..., n) thenP
cyclic
(a1+a2)2(a1+a3)
2
a1(a21+a2a3)≥ 8
nPk=1
ak.
Mihaly Bencze
PP29363. If ak > 0 (k = 1, 2, ..., n) thenQ
cyclic
�a21 + a2a3
�≤
�
cyclic
(a1+a2)3
4nn�
k=1ak
.
Mihaly Bencze
PP29364. In all triangle ABC holds
1).Q �
a2 + bc�≤ s2(s2+r2+2Rr)
3
32Rr
2).Q �
r2a + rbrc�≤ s4R3
r
3).Q �
sin4 A2 + sin2 B
2 sin2 C2
�≤ ((2R−r)(s2+r2−8Rr)−2Rr2)
3
1048576R8(2R−r)
4).Q �
cos4 A2 + cos2 B
2 cos2 C2
�≤ ((4R+r)3+s2(2R+r))
3
1048576R8(4R+r)
Mihaly Bencze
984 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP29365. If x ∈ R thensin6 x cos6 x
sin8 x+cos6 x+cos2 x+1+ 8 cos6 x
cos8 x+sin2 x+9+ 8 sin6 x
sin6 x+16≤ sin8 x+cos8 x+6 sin2 x cos2 x+16
6 .
Mihaly Bencze
PP29366. If ak > 0 (k = 1, 2, ..., n) thenP
cyclic
�
a1(a21+a1a2)(a2+a3)
a1+a3≤
nPk=1
ak.
Mihaly Bencze
PP29367. If x ∈�0, π2
�then
sin2 x√1 + cos2 x+ cos2 x
p1 + sin2 x ≤ 1 + sin3 x+ cos3 x.
Mihaly Bencze
PP29368. If x, y > 0 then
xy�xp2 (x2 + y2) + y2
��yp
2 (x2 + y2) + x2�≤
�2x3 + y3
� �2y3 + x3
�.
Mihaly Bencze
PP29369. If a, b, c > 0 thenP
a4 +P b5c3
a4+P a8
bc3≥ 9
��
a3�
a
�2.
Mihaly Bencze
PP29370. If a, b, c > 0, a 6= b 6= c thenP a2(an−bn)
(a−b)bn−1 ≥ 3(n+1)�
a3�
a for all
n ∈ N∗.
Mihaly Bencze
PP29371. If a > 0, 0 < b ≤ xk ≤ c (k = 1, 2, ..., n) , thenP
cyclic
x1
x21+a2x2x3
≤ n2(b+c)2
8abcn�
k=1xk
.
Mihaly Bencze
PP29372. In all triangle ABC holdsPa2b2 cos2C ≤ 3
�s2 − r2 − 4Rr
�2 − 16s2Rr.
Mihaly Bencze
Proposed Problems 985
PP29373. In all triangle ABC holds
1).P
a2√b4 + c4 ≤ 2
√2��
s2 + r2 + 4Rr�2 − 20s2Rr
�
2).P
r2a
qr4b + r4c ≤
√2�2s4 − 5s2r (4R+ r)
�
3).P
sin4 A2
qsin8 B
2 + sin8 C2 ≤
√2
�2�s2+r2−8Rr
16R2
�2− 5r2(2R−r)
32R3
�
4).P
cos4 A2
qcos8 B
2 + cos8 C2 ≤
√2
�(s2+(4R+r)2)
2
128R4 − 5s2(4R+r)32R3
�
Mihaly Bencze
PP29374. If ak > 0 (k = 1, 2, ..., n) thenP
cyclic
r3a1a2 +
qa41+a42
2 ≤ 2nP
k=1
ak.
Mihaly Bencze
PP29375. If a, b, c > 0 thenP
a2 ≥P ab+ 14
P (a2−b2)2
a2+b2.
Mihaly Bencze
PP29376. If ak > 0 (k = 1, 2, ..., n) then
n−12
nPk=1
a2k −P
1≤i<jnaiaj ≥ 1
4
P1≤i<jn
(a2i−a2j)2
a2i+a2j.
Mihaly Bencze
PP29377. In all triangle ABC holds
3P �
tg6A2 + tg6B2�tg2C2 ≥ 12r2(4R+r)2
s4+ 2
√3r2
s2.
Mihaly Bencze
PP29378. If a, b, c > 0, thenP
a2 + 2P a2b2
a2+b2≥ 2
Pab.
Mihaly Bencze
PP29379. If a, b, c, x, y > 0 thenP a
b+xc +y�
ab�
a2≥ 2
�1
x+1 +q
yx+1
�.
Mihaly Bencze
PP29380. If x, y, z > 0 and xyz = 1 thenP 1
(x+1)2+(y+1)2−2≤ 1
2 .
Mihaly Bencze
986 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP29381. If a, b, c > 0 thenP
a2√b4 + c4 ≤ 2
√2P
a2b2 −√2abc
Pa.
Mihaly Bencze
PP29382. Determine all a, b ∈ R for which√a+
√b+
p2 (a+ b) ≥ 3 +
4(3−√6)
3
�ab+ 2 (a+ b)2 − 3
�.
Mihaly Bencze
PP29383. If x, y, z ∈ R then1).
P sin2 x3+sin2 y+sin2 z
+ cos2 x cos2 y cos2 z ≤ 1
2).P cos2 x
3+cos2 y+cos2 z+ sin2 x sin2 y sin2 z ≤ 1
Mihaly Bencze
PP29384. In all triangle ABC holds
1). (2s)32�√
6s−P√
a�≤ 4
�√3−
√2� �
s2 − 3r2 − 12Rr�
2). s32
�√3s−P
√s− a
�≤ 4
�√3−
√2� �
s2 − 3r2 − 12Rr�
3). (4R+ r)32
�p3 (4R+ r)−P√
ra
�≤ 4
�√3−
√2� �
(4R+ r)2 − 3s2�
4).�2R−r2R
� 32
�q3(2R−r)
2R −P sin A2
�≤ 4
�√3−
√2� � (4R+r)2−3s2
16R2
�
5).�4R+r2R
� 32
�q3(4R+r)
2R −P cos A2
�≤ 4
�√3−
√2� � (4R+r)2−3s2
16R2
�
Mihaly Bencze
PP29385. If a, b, c > 0 then
(P
a)32
�p3P
a−P√
a�≤ 4
�√3−
√2� �P
a2 −P
ab�.
Mihaly Bencze
PP29386. If x ∈�0, π2
�then
1). sinx+ cosx+√2−
√3 ≥ 4(3−
√6)
3
�sin2 x cos2 x− 1
�
2).√2 (sinx+ cosx)− 2 ≥ 4(3−
√6)
3
�sin2 2x− 1
�
Mihaly Bencze
Proposed Problems 987
PP29387. If a, b, c, x, y > 0 and a+ b+ c = 1 thenPcyclic
an+2
(b+x)(c+y) ≥1
3n−1(3x+1)(3y+1).
Mihaly Bencze
PP29388. In all triangle ABC holdsPq
tgA2 tg
B2 ≥
√3 + 4
�√3−
√2� �3r(4R+r)
s2− 1
�.
Mihaly Bencze
PP29389. In all triangle ABC holds
1).P √
3+7 sinAsinA ≥ 147
2√6− s
2√6R
2).P √
3+7 sin2 A
sin2 A≥ 49
2√6
�1− s2−(2R+r)2
2R2
�
Mihaly Bencze
PP29390. In all acute triangle ABC holds
1).P √
3+7 cosAcosA ≥ 174
2√6− R+r
2√6R
2).P √
3+7 cos2 Acos2 A
≥ 492√6
�s2−r(4R+r)
2R2
�
Mihaly Bencze
PP29391. In all triangle ABC holds
1).P
�
3+7 sin2 A2
sin2 A2
≥ 494√6
�4R+rR
�
2).P
�
3+7 cos2 A2
cos2 A2
≥ 494√6
�2R−rR
�
Mihaly Bencze
PP29392. In all triangle ABC holds
1).Pp
ra (7r + 3ra) ≥ 49r(2R−r)√6
2).Pp
ha (7r + 3ha) ≥49r(s2+r2−2Rr)
4√6R
Mihaly Bencze
PP29393. If a, b, c > 0 thenP a(b+λc)2
b2+bc+c2≥ (λ+1)2
�
ab�
a for all λ ≥ 0.
Mihaly Bencze
988 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP29394. If a, b, c > 0 then
1). (a+b)3
a2+ab+b2+ 2(a+b)
3 ≥ 2a(a+2b)2a+b + 2b(b+2a)
2b+a
2).�
a(a+b)2
a2+ab+b2+ 2b
3
��b(a+b)2
a2+ab+b2+ 2a
3
�≥ 4ab
Mihaly Bencze
PP29395. If ak > 0 (k = 1, 2, ..., n) , then�nP
k=1
aλk
��nP
k=1
1aλk
�≥ n 3
s3n
�nP
k=1
ak
��nP
k=1
1ak
�− 2n3 for all λ ≥ 1
Mihaly Bencze
PP29396. If a, b, c > 0 and abc = 1 thenP a2(1+a2+a4)
b(1−b+b2)≥ 9
P 1a+b2+c
.
Mihaly Bencze
PP29397. In all triangle ABC holds
1).P 1√
a2+ab+b2≥ 1√
s2+r2+4Rr
�2 +
(s2+r2+4Rr)2+8s2Rr
2s2(s2+r2+2Rr)
�
2).P 1√
r2a+rarb+r2b
≥ 1s
�2 + s2+r(4R+r)
2R(4R+r)
�
Mihaly Bencze
PP29398. In all triangle ABC holds
1).P 1√
h2a+hahb+h2
b
≥ 2s
qR2r
�1 +
2Rr(5s2+r2+4Rr)(s2+r2+4Rr)(s2+r2+2Rr)
�
2).P 1√
(s−a)2+(s−a)(s−b)+(s−b)2≥ 1√
r(4R+r)
�2 +
r(s2+(4R+r)2)2s2R
�
Mihaly Bencze
PP29399. If a, b > 0 then√2ab+ a2
�2√
a2+ab+b2+ 1√
3a
�+√2ab+ b2
�2√
a2+ab+b2+ 1√
3b
�≥
≥ 4 + 22a+b
�92 + 2ab
a+b
�+ 2
2b+a
�92 + 2ab
a+b
�.
Mihaly Bencze
PP29400. If ak > 0 (k = 1, 2, ..., n) andnP
k=1
a2k ≤ 2n thennP
k=1
qa3k + 1 ≤ 2n.
Mihaly Bencze
Proposed Problems 989
PP29401. If ak > 0 (k = 1, 2, ..., n) andnP
k=1
qa3k + 1 ≥ 3n then
nPk=1
a2k ≥ 4n.
Mihaly Bencze
PP29402. In all triangle ABC holds
1). 4− r(4R+r)s2
≥Pq
1 + 8r3
h3a
2). 5− 4r(4R+r)s2
≥Pq1 + 8r3
r3a
Mihaly Bencze
PP29403. In all triangle ABC holdsPp
3√3 + 64 sin3A ≤
q√33
�4(s2−r(4R+r))
3R2 + 3
�.
Mihaly Bencze
PP29404. In all acute triangle ABC holdsP√
1 + 64 cos3A ≤ 11− 4(s2−(2R+r)2)R2 .
Mihaly Bencze
PP29405. In all triangle ABC holds
1).Pq
1 + 512 sin6 A2 ≤ 35 +
4(r2−s2)R2
2).Pq
1 + 51227 cos6 A
2 ≤ 3 +4((4R+r)2−s2)
9R2
Mihaly Bencze
PP29406. Solve in (0,+∞) the following system:
�a21 + 2
� �a22 + 2
�= 4q�
a33 + 1� �
a34 + 1�
�a22 + 2
� �a23 + 2
�= 4q�
a34 + 1� �
a35 + 1�
−−−−−−−−−−−−−−−−−−�a2n + 2
� �a21 + 2
�= 4q�
a32 + 1� �
a33 + 1�
.
Mihaly Bencze
990 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP29407. If ak > 0 (k = 1, 2, ..., n) , then 2nλ2 +nP
k=1
a2k ≥ 2nP
k=1
qλa3k + λ4
for all λ ≥ 0.
Mihaly Bencze
PP29408. If ak > 0 (k = 1, 2, ..., n) andnP
k=1
1√a3k+λ3
≤ 1 then
nPk=1
a2k ≥ 2n√λ�n−
√λ�for λ > 0.
Mihaly Bencze
PP29409. If ak > 0 (k = 1, 2, ..., n) then
2nP
k=1
a3k+1
a2k+2
≤nP
k=1
qa3k + 1 ≤ 1
2
nPk=1
a2k + n.
Mihaly Bencze
PP29410. If x ∈�0, π2
�, then
2 sin2 x cos2 x+ (sinx+ cosx) (1− sinx cosx) +
2�p
1 + sin5 x+√1 + cos5 x
�≤ 6.
Mihaly Bencze
PP29411. If ak > 0 (k = 1, 2, ..., n) , thennP
k=1
ra3k + 2
qa5k + 1− 7
4 ≤ n2 +
nPk=1
a2k.
Mihaly Bencze
PP29412. If x, y, z > 0, thenP
cyclic
x4λ−2√2(y4+z4)+yz
≥ 19λ−1
�Px2
�2λ−2for all
λ ≥ 1.
Mihaly Bencze
PP29413. If xk > 0 (k = 1, 2, ..., n) thenP
cyclic
xn−11
�
(n−1)(x2n−22 +x2n−2
3 +...+x2n−2n )+x2x3...xn
≥ 1.
Mihaly Bencze
Proposed Problems 991
PP29414. If 0 < λ ≤ ak (k = 1, 2, ..., n) then
λnP
k=1
ak ≤P
a1a2 ≤ 1λ
Pa1a2a3.
Mihaly Bencze
PP29415. If 0 < λ ≤ ak (k = 1, 2, ..., n) , then
λn
�nP
k=1
ak
�n
≤ Qcyclic
(λ+ a1 + a2 + ...+ an−1)nQ
k=1
ak.
Mihaly Bencze
PP29416. If n,m, k ∈ N ∗ then��
1 + 1n
�n+�1 + 1
m
�m+�1 + 1
k
�k�3≤
��1 + 1
n
�n �1 + 1
m
�m+�1 + 1
k
�k���1 + 1
m
�m �1 + 1
k
�k+�1 + 1
n
�n� ·
·��
1 + 1k
�k �1 + 1
n
�n+�1 + 1
m
�m�.
Mihaly Bencze
PP29417. If ak > 0 (k = 1, 2, ..., n) , then�
nPk=1
ak
��nP
k=1
1ak
�≥ n2 + 2
n(n−1)
P
1≤i<j≤≤n
|ai−aj |√aiaj
!2
.
Mihaly Bencze
PP29418. If x, y, z > 0 and x 6= y 6= z thenP 1
(x−y)2+ 1
4
P 1xy ≥ 9
�
xy .
Mihaly Bencze
PP29419. If x, y, z > 0 thenPq
xy+z ≤
r2 (P
xy)�P 1
(x+y)2
�.
Mihaly Bencze
PP29420. If xk > 0 (k = 1, 2, ..., n), then P
1≤i<j≤nxixj
! P
1≤i<j≤n
1(xi+xj)
2
!≥
�n(n−1)
2
�2.
Mihaly Bencze
PP29421. If a, b, c > 0, then (a+ b+ c)15 ≥ 14348907a4b4c4�a3 + b3 + c3
�.
Mihaly Bencze
992 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP29422. In all triangle ABC holds1).
P 1r3a
≤ s8
4782969r11
2).P 1
h3a≤ 16s8
4782969r8R4
Mihaly Bencze
PP29423. In all triangle ABC holds r8 ≥ 4782969s6�s2 − 12Rr
�.
Mihaly Bencze
PP29424. If ak > 0 (k = 1, 2, ..., n) andnP
k=1
ak = n then
�nP
k=1
ank
��nQ
k=1
ak
�n+1
≤ n.
Mihaly Bencze
PP29425. If xk > 0 (k = 1, 2, ..., n) then
Pcyclic
x1x2
+2n+1
n�
k=1
xk
(x1+x2)(x2+x3)...(xn+x1)≥ n+ 2.
Mihaly Bencze
PP29426. If xk > 0 (k = 1, 2, ..., n) thenP x1
x2+x3+ 2n+1
�
(x1+x2)�
(x1+x2+x3)≥ n+ 2.
Mihaly Bencze
PP29427. If ak > 0 (k = 1, 2, ..., n) then
Pcyclic
r(a21+a22)(a1+a2)
2
a1a2+ 4a1a2 ≥ 2
√3
nPk=1
ak.
Mihaly Bencze
PP29428. If ak > 0 (k = 1, 2, ..., n) and λ > 0 then
12
nPk=1
ak +2n
(a1+λ)(a2+λ)...(an+λ) ≥ n− nλ2 + 1.
Mihaly Bencze
Proposed Problems 993
PP29429. In all triangle ABC holdsPs
�
ctgA2
�
ctgB2+�
ctgC2
≥q
32
PqtgA
2 tgB2 .
Mihaly Bencze
PP29430. If xk > 0 (k = 1, 2, ..., n) then
P x1x2
+2n
n�
k=1xk
(x1+x2)(x2+x3)...(xn+x1)≥ n+ 1.
Mihaly Bencze
PP29431. If a, b > 0 then a2+b2
a2+ab+b2+ 10
3 ≥ 4a(a+2b)(a+b)(2a+b) +
4b(b+2a)(a+b)(2b+a) .
Mihaly Bencze
PP29432. In all triangle ABC holds
1).P 1
�
ra(√rb+
√rc)
≥q
3√r
2s
P 1√ra
2).P 1
�
ha(√hb+
√hc)
≥q
3√R
2s√2
P 1√ha
Mihaly Bencze
PP29433. If ak > 0 (k = 1, 2, ..., n) thenP
cyclic
��a31+a322a1a2
�2+ a1a2
�≥ 2
nPk=1
a2k.
Mihaly Bencze
PP29434. If ak > 0 (k = 1, 2, ..., n) thenP
cyclic
r�a31+a322a1a2
�2+ a1a2 ≥
√2
nPk=1
ak.
Mihaly Bencze
PP29435. If ak > 0 (k = 1, 2, ..., n) thenP
cyclic
(a31+a32)2
a41+a21a22+a42
≥ 4(�
a1a2)2
2n�
k=1a2k+�
a1a2
.
Mihaly Bencze
PP29436. If a, b, c > 0, thenP ((a3+b3)c)
2
a4+a2b2+b4≥ 36a2b2c2
2�
a2+�
ab.
Mihaly Bencze
994 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP29437. If ak > 0 (k = 1, 2, ..., n) thenP
cyclic
pa21 − a1a2 + a22 ≥
nPk=1
ak.
Mihaly Bencze
PP29438. If ak > 0 (k = 1, 2, ..., n) thenP
cyclic
pa21 − a1a2 + a22 ≤
P a21a2.
Mihaly Bencze
PP29439. Prove thatnP
k=1
3
r4(k2+k+1)
(2k3+3k2+3k+1)2≤ n
n+1 .
Mihaly Bencze
PP29440. If a, b, c > 0 and�
1ab
�3+�1bc
�3+�
1ca
�3= 4 then
P a2−ab+b2
(a3+b3)2≤ 1.
Mihaly Bencze
PP29441. If a, b, c > 0 thenP a4(b+c)√
(a2+ab+b2)(a2+ac+c2)≥ 1
3
Pa2 (b+ c) .
Mihaly Bencze
PP29442. If a, b, c > 0 thenP a2
b ≥P√a2 − ab+ b2.
Mihaly Bencze
PP29443. If a, b > 0 then 2q
2a4+b4
+ 12
�1a2
+ 1b2
�+ 2
�a2
b + b2
a
�≥ 7.
Mihaly Bencze
PP29444. If ak > 0 (k = 1, 2, ..., n) then
Pcyclic
�a21a2
+a22a1
�pa21 + a1a2 + a22 ≥ 2
Ppa41 + a21a
22 + a42.
Mihaly Bencze
PP29445. If a, b > 0 then�2a3
a3+b3+ b3
2a3
��2b3
a3+b3+ a3
2b3
�≥ (2a+b)3(a+2b)3
36ab(2a2+b2)(a2+2b2).
Mihaly Bencze
Proposed Problems 995
PP29446. If a, b > 0 then 9�
2a3
a3+b3+ b3
2a3− 5
6
��2b3
a3+b3+ a3
2b3− 5
6
�≥
≥�a(a+2b)2a2+b2
+ 2a2+b2
a(a+2b)
��b(b+2a)2b2+a2
+ 2b2+a2
b(b+2a)
�.
Mihaly Bencze
PP29447. In all triangle ABC holds1).
P a3
b3+c3≥ 2s2
s2+r2+4Rr
2).P r3a
r3b+r3c
≥ (4R+r)2
2s2
3).P (sin A
2 )6
(sin B2 )
6+(sin C
2 )6 ≥ 2(2R−r)2
s2+r2−8Rr
4).P (cos A
2 )6
(cos B2 )
6+(cos C
2 )6 ≥ 4(4R+r)2
s2+(4R+r)2
Mihaly Bencze
PP29448. If a, b > 0 then
�2a√3+ b2√
a2+ab+b2
��2b√3+ a2√
a2+ab+b2
�≥ a2 + ab+ b2.
Mihaly Bencze
PP29449. In all triangle ABC holds 4R+rs +
√3�
1
3n2 +1
P �ctgA
2
�n� 1k ≥ 2
√3
for all n, k ∈ N, n ≥ 2, k ≥ 6.
Mihaly Bencze
PP29450. If ak > 0 (k = 1, 2, ..., n) , thenPcyclic
a21�
(a21+a1a2+a22)(a21+a1a3+a23)≥ n
3 .
Mihaly Bencze
PP29451. Prove thatnP
k=1
�3k2+6k+53k2+6k+2
+ 4k(k+2)(2k+1)(2k+3)
�≥ 2m.
Mihaly Bencze
PP29452. If a, b > 0 then a2+ab+b2
a+√ab+b
+ 8ab√ab
(a+b)(√a+
√b)
2 ≥ 2√ab.
Mihaly Bencze
996 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP29453. If x ∈�0, π2
�then
5(1+sinx)(1+cos x)−1 + 8 sin 2x
(2+sinx)(2+cosx)(sinx+cosx) ≥ 2.
Mihaly Bencze
PP29454. If x ∈ R, then 1−sin2 x cos2 x1+sin2 x cos2 x
+ sin2 2x
(1+sin2 x)(1+cos2 x)≥ 1.
Mihaly Bencze
PP29455. If a, b, c > 0 then 1 +P (a−b)(a−c)
b+c ≥ 2P a
√bc
b+c .
Mihaly Bencze
PP29456. If a, b > 0 then a2+2b2
b(2a+b) +b2+2a2
a(2b+a) +8ab
(a+b)2≥ 4.
Mihaly Bencze
PP29457. If a, b, c > 0 then 1 +P (a−b)(a−c)
b+c ≤ 12√abc
Pa2√a+ 1
2
P√ab.
Mihaly Bencze
PP29458. If a, b, c > 0 then 1 +P (a−b)(a−c)
b+c ≥Pq
(a2+bc)(b2+ca)(b+c)(c+a) .
Mihaly Bencze
PP29459. If xk > 0 (k = 1, 2, ..., n) andnQ
k=1
(xk + 1) = n then
Pcyclic
x41+3x3
1+5x21+7x1+4
x32+2x2
2+3x2+4≥ n n
√n.
Mihaly Bencze
PP29460. If a, b, c > 0 then 1 +P (a−b)(a−c)
b+c ≤ 2P
a+ 2P a2−bc
b+c .
Mihaly Bencze
PP29461. If x, y > 0 and xy + x+ y = 1 then
4x3+7x2+5x+24y2+3y+2
+ 4y3+7y2+5y+24x2+3x+2
≥ 2√2.
Mihaly Bencze
Proposed Problems 997
PP29462. If x, y, z > 0 and xyz + xy + yz + zx+ x+ y + z = 2 then5x3+6x2+5x+4
5y2+y+4+ 5y3+6y2+5y+4
5z2+z+4+ 5z3+6z2+5z+4
5x2+x+4≥ 3 3
√3.
Mihaly Bencze
PP29463. If a, b, c > 0 thenP a
b2+bc+c2≥
�
a�
ab .
Mihaly Bencze
PP29464. If a, b, c > 0 thenP a
7bc+ 3√
4(b6+c6)≥
�
a3�
ab .
Mihaly Bencze
PP29465. If a, b, c > 0 thenP ab2
7bc+ 3√
4(b6+c6)≥
�
ab3�
a .
Mihaly Bencze
PP29466. If ak > 0 (k = 1, 2, ..., n) thenP
cyclic
r10a1a2 +
3
q4�a61 + a62
�≤ 2
√3
nPk=1
ak.
Mihaly Bencze
PP29467. If ak > 0 (k = 1, 2, ..., n) , thenP
cyclic
r3
q4�a61 + a62
�− 2a1a2 ≤
√3P |a1 − a2| .
Mihaly Bencze
PP29468. If ak > 0 (k = 1, 2, ..., n) , thenP
cyclic
|a31−a32|7a1a2+
3�
4(a61+a62)≥ 1
3
Pcyclic
|a1 − a2| .
Mihaly Bencze
PP29469. If ak > 0 (k = 1, 2, ..., n) then
Pcyclic
a31−a32
7a1a2+3�
4(a61+a62)
!2
≥ 29
nPk=1
a2k − 29
Pa1a2.
Mihaly Bencze
998 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP29470. If ak > 0 (k = 1, 2, ..., n) thenP
cyclic
a21+a1a2+a22
7a2a3+3�
4(a62+a63)≥ n
3 .
Mihaly Bencze
PP29471. If ak > 0 (k = 1, 2, ..., n) then
3P
cyclic
��a31 − a32�� ≥ P
cyclic
|a1 − a2|�7a1a2 +
3
q4�a61 + a62
��.
Mihaly Bencze
PP29472. If x ∈ R thensin2 x
3√
4(64+cos12 x)+14 cos2 x+ cos2 x
3�
4(64+sin12 x)+14 sin2 x+ (sinx)
23+(cosx)
23
12 ≥ 7− 3√212 .
Mihaly Bencze
PP29473. If ak > 0 (k = 1, 2, ..., n) thenP
cyclic
a31+a32
a1a2+3�
4(a61+a62)≥ 2
3
nPk=1
ak.
Mihaly Bencze
PP29474. If a, b, c > 0 then√3 (P
a)112 ≥ 243 (
Pab)3 .
Mihaly Bencze
PP29475. If a, b, c > 0 thenP a4
b4�
18ab−(a+b+c)√
3(a+b+c)c+2(a+b+c)2� ≥ 3
�
a2
(�
a)4.
Mihaly Bencze
PP29476. If x ∈�0, π2
�then
tg8x
2−√2+2 sin2 x cos2 x
+ 16sin8 x(4 sin2 x+2−cosx)
+ cos8 x16(4 cos2 x+2−sinx)
≥ 5−2 sin2 x cos2 x3 .
Mihaly Bencze
PP29477. If x, y, z > 0 such that xn + yn + zn = 3. Determine all n, k ∈ Nfor which (xy)n+k + (yz)n+k + (zx)n+k ≤ 3.
Mihaly Bencze
PP29478. If x ∈�0, π2
�then (sinx cosx)
52 + 2 4
√2�(sinx)
52 + (cosx)
52
�≤ 3.
Mihaly Bencze
Proposed Problems 999
PP29479. If x ∈�0, π2
�then
3�√
sinx+√cosx+ 4
√2�2
≥�2 + sin2 x cos2 x
�3.
Mihaly Bencze
PP29480. In all triangle ABC holds
�√23
PqctgA
2 ctgB2
�2
≥P�q
tgA2 ctg
B2 +
qtgA
2 ctgC2
�.
Mihaly Bencze
PP29481. If x ∈�0, π2
�then
23
�1
sinx + 1cosx + 1√
2
�2≥ (sinx+ cosx)
�1 +
√2
sinx cosx
�+ tgx+ ctgx.
Mihaly Bencze
PP29482. If x ∈�0, π2
�then
�sinx
√sinx
cos2 x√cosx
+ cosx√cosx
2 4√2+
4√8sin2 x
√sinx
�·
·�
cosx√cosx
sin2 x√sinx
+ sinx√sinx
2 4√2+
4√8cos2 x
√cosx
�≥ 9.
Mihaly Bencze
PP29483. If a, b > 0 then�2 4√a+
4√b
a(a+2b)
�2
+�2
4√b+ 4√a
b(b+2a)
�2
≥ a2+4ab+b2
3 .
Mihaly Bencze
PP29484. In all triangle ABC holds
1).P√
ra ≥q
9√3
2s
Pra
�√rb +
√rc�
2).P√
ha ≥q
9√Rr
2s√2
Pha
�√hb +
√hc�
Mihaly Bencze
PP29485. If xk ≥p√
2− 1 (k = 1, 2, ..., n) and λ0 =1+√√
2−1
1+4√√
2−1then
nPk=1
1+xk
1+√xk
≥ nλ0.
Mihaly Bencze
1000 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP29486. In all triangle ABC holdsP
4
qtgA
2 tgB2 ≥ 9r(4R+r)
s3
q√3r (4R+ r).
Mihaly Bencze
PP29487. If ak > 0 (k = 1, 2, ..., n) andnP
k=1
a2k = n, then
nPk=1
1ak
≥r
nn−1
Pcyclic
a1+a2+...+an−1
an.
Mihaly Bencze
PP29488. If ai ≥ 1 (i = 1, 2, ..., n) , then
λ+ k
snQ
i=1ai
!nQ
i=1
λ+aiλ+ k
√ai
≥ λ+ 1
for all k ≥ 1, k ∈ N and for all λ > 0.
Mihaly Bencze
PP29489. If ak > 0 (k = 1, 2, ..., n) andnP
k=1
apk = n (p ≥ 1) then P
cyclic
ar1ar+s2
! P
cyclic
ar2ar+s1
!≥ n2 for all r, s > 0.
Mihaly Bencze
PP29490. If ak > 0 (k = 1, 2, ..., n) andnP
k=1
a3k = n thenP
cyclic
1a21(a22−a2a3+a23)
≥ n.
Mihaly Bencze
PP29491. If ai,λi > 0 (i = 1, 2, ..., n) andnP
i=1λi = 1 then
�aλ11 aλ2
2 ...aλnn
�(k+1)n+�aλ21 aλ3
2 ...aλ1n
�(k+1)n+ ...+
�aλn1 aλ1
2 ...aλn−1n
�(k+1)n≥
≥ anλ1+k1 anλ2+k
2 ...anλn+kn + anλ2+k
1 anλ3+k2 ...anλ1+k
n + ...
+anλn+k1 anλ1+k
2 ...anλn−1+kn .
Mihaly Bencze
Proposed Problems 1001
PP29492. If a, b > 0 thenq
9 + 3(2a2+b2)(2b2+a2)a2b2
≥ 4 + ab +
ba .
Mihaly Bencze
PP29493. If aij > 0 (i = 1, 2, ...,m; j = 1, 2, ..., n) then�mPi=1
1ai1ai2 ...ain
�mPi=1
(ai1 + ai2 + ain)n ≥ m2nn.
Mihaly Bencze
PP29494. If a, b > 0 then 1 + ab +
ba ≥ (2a+b)(a+2b)
3ab .
Mihaly Bencze
PP29495. If ak > 0 (k = 1, 2, ..., n) thensn2(n−1)(n−2)
2 + n(n−1)2
�nP
k=1
a2k
��nP
k=1
1a2k
�≥ P
cyclic
a1+a2+...+an−1
an.
Mihaly Bencze
PP29496. If a, b > 0 then�1 +
√(2a2+b2)(2b2+a2)
ab
�2
≤�q
(2a+b)(a+2b)ab − 1
�2
.
Mihaly Bencze
PP29497. If ak > 0 (k = 1, 2, ..., n) then
�nP
k=1
ak
��nP
k=1
1ak
�≤ n
P a1a2.
Mihaly Bencze
PP29498. If xi > 0 (i = 1, 2, ..., n) and k ∈ {2, 3, ..., n} thenP x1x2+x3+...+xk
+2n+1
�
(x1+x2+...+xk−1)�
(x1+x2+...+xk)≥ n+ 2.
Mihaly Bencze
PP29499. If xi > 0 (i = 1, 2, ..., k) when k = n (n− 2) then
(n− 2)
1 +
vuut1 +
s�kP
i=1a2i
��kP
i=1
1a2i
� ≤
s�kP
i=1ai
��kP
i=1
1ai
�.
Mihaly Bencze
1002 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP29500. In all triangle ABC holds
1).P� √
a+√b
a+√ab+b
�2≥ 4(5s2+r2+4Rr)
9s(s2+r2+4Rr)
2).P� √
s−a+√s−b
c+√
(s−a)(s−b)
�2
≥ 2(s2+r2+4Rr)9sRr
3).P� √
ra+√rb
ra+√rarb+rb
�2≥ 2((4R+r)2+s2)
9s2R
4).P� √
ha+√hb
ha+√hahb+hb
�2≥
2�
(s2+r2+4Rr)2+8s2Rr
�
9s2r(s2+r2+2Rr)
5).P�
sin A2+sin B
2
sin2 A2+sin A
2sin B
2+sin2 B
2
�2
≥ 16R(16R2−24Rr+5r2+s2)9((2R−r)(s2+r2−8Rr)−2Rr2)
6).P�
cos A2+cos B
2
cos2 A2+cos A
2cos B
2+cos2 B
2
�2
≥ 16R(5(4R+r)2+s2)9((4R+r)3+s2(2R+r))
Mihaly Bencze
PP29501. In all triangle ABC holds
1).P
(b+ c)�a+
√ab+b√
a+√b
�2≤ 9
8
�5s2 + r2 + 4Rr
�
2).P
a
�c+√
(s−a)(s−b)√s−a+
√s−b
�2
≤ 98
�s2 + r2 + 4Rr
�
3).P
(rb + rc)�ra+
√rarb+rb√
ra+√rb
�2≤ 9
8
�(4R+ r)2 + s2
�
4).P
(hb + hc)�ha+
√hahb+hb√
ha+√hb
�2≤
9�
(s2+r2+4Rr)2+8s2Rr
�
32R2
5).P �
sin2 B2 + sin2 C
2
�� sin2 A2+sin A
2sin B
2+sin2 B
2
sin A2+sin B
2
�2
≤ 9(16R2−24Rr+5r2+s2)128R2
6).P �
cos2 B2 + cos2 C
2
�� cos2 A2+cos A
2cos B
2+cos2 B
2
cos A2+cos B
2
�2
≤ 9(5(4R+r)2+s2)128R2
Mihaly Bencze
PP29502. In all triangle ABC holds
1).P�
a+√ab+b√
a+√b
�2≤ 9s
2
2).P�
c+√
(s−a)(s−b)√s−a+
√s−b
�2
≤ 9s4
3).P�
ra+√rarb+rb√
ra+√rb
�2≤ 9(4R+r)
4
4).P�
ha+√hahb+hb√
ha+√hb
�2≤ 9(s2+r2+4Rr)
8R
5).P�
sin2 A2+sin A
2sin B
2+sin2 B
2
sin A2+sin B
2
�2
≤ 9(2R−r)8R
Proposed Problems 1003
6).P�
cos2 A2+cos A
2cos B
2+cos2 B
2
cos A2+cos B
2
�2
≤ 9(4R+r)8R
Mihaly Bencze
PP29503. If a, b, c > 0 and a 6= b 6= c thenP�
|a−b|+√
|ab+b−ca−b2|+|b−c|√|a−b|+
√|b−c|
�2
≤ 92 (max {a, b, c}−min {a, b, c}) .
Mihaly Bencze
PP29504. Prove thatnP
k=0
�nk
� ��nk
�+q�
nk
�+ 1
�2
≤ 92
�2n +
�2nn
��.
Mihaly Bencze
PP29505. Prove thatnP
k=1
�13k2−3k−3+
√36k4−12k3−17k2+3k+2
(√4k2−1+
√9k2−3k−2)
√36k4−12k3−17k2+3k+2
�2
≤ 9(3n2+3n+1)8(2n+1)(3n+1) .
Mihaly Bencze
PP29506. Prove thatnP
k=1
�√k!((k+1)2+
√k2+k+1)√
k+√k2+k+1
�2
≤ 98 ((n+ 2)!− 1) .
Mihaly Bencze
PP29507. Prove that6481
nPk=1
�k2+k+
√k2+k+1√
k2+k+1
�2 �k2−k+
√k2−k+1√
k2−k+1
�2≤ n+
n(n+1)(2n+1)(3n2+3n+4)30 .
Mihaly Bencze
PP29508. Denote Fk and Lk the kth Fibonacci respective Lucas numbers.Prove that
1). 9 (Fn+2 + Ln+2 − 4) ≥ 8nP
k=1
�Fk+
√FkLk+Lk√
Fk+√Lk
�2
2). 9 (FnFn+1 + LnLn+1 − 9) ≥ 8nP
k=1
�F 2k+FkLk+L2
k
Fk+Lk
�2
Mihaly Bencze
1004 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP29509. Prove thatnP
k=1
�2k+1+
√k(k+1)√
k+√k+1
�2
≤ 9n (n+ 2) .
Mihaly Bencze
PP29510. Prove thatnP
k=1
� √k+1√
k(k+√k+1)
�2
≥ 8n9(n+1) .
Mihaly Bencze
PP29511. Solve in�0, π2
�the following system:
8 (1 + sinx1 cosx1)2 = 9 (1 + sin 2x2)
8 (1 + sinx2 cosx2)2 = 9 (1 + sin 2x3)
−−−−−−−−−−−−−−−−8 (1 + sinxn cosxn)
2 = 9 (1 + sin 2x1)
.
Mihaly Bencze
PP29512. If ak > 0 (k = 1, 2, ..., n) , then 9nP
k=1
ak ≥ 4P
cyclic
�a1+
√a1a2+a2√
a1+√a2
�2.
Mihaly Bencze
PP29513. If ak > 0 (k = 1, 2, ..., n) , then
9
�nP
k=1
ak
� nP
k=1
ak +P
cyclic
√a1a2
!≥ 2
2
nPk=1
ak +P
cyclic
√a1a2
!2
.
Mihaly Bencze
PP29514. Solve in R the following system:
8�a1 +
√a1a2 + a2
�2= 9 (a2 + a3)
�√a3 +
√a4�2
8�a2 +
√a2a3 + a3
�2= 9 (a3 + a4)
�√a4 +
√a5�2
−−−−−−−−−−−−−−−−−−−−−−8�an +
√ana1 + a1
�2= 9 (a1 + a2)
�√a2 +
√a3�2
.
Mihaly Bencze
PP29515. If a, b > 0 then (a+2b)(b+2a)
(a+b)4≤ 9
16ab .
Mihaly Bencze
Proposed Problems 1005
PP29516. If a, b > 0 then 13√a2b
+ 13√ab2
≥ 4a+b .
Mihaly Bencze
PP29517. If ak > 0 (k = 1, 2, ..., n) , thenP
cyclic
1(a1+2a2)(a2+2a1)
≥ n3
9
�
n�
k=1
ak
�2 .
Mihaly Bencze
PP29518. If a, b, c > 0 thenP a
(a+b)(a+c) ≤9
4�
a .
Mihaly Bencze
PP29519. If ak > 0 (k = 1, 2, ..., n) , thenP an−1
1(a1+a2)(a1+a3)...(a1+an)
≤ n2
2n−1n�
k=1ak
.
Mihaly Bencze
PP29520. If a, b, c > 0 and λ > 0 thenP 1
(λa2+bc)n≥ 3n+1
(λ+1)n(�
a2)nfor all
n ∈ N.
Mihaly Bencze
PP29521. If a, b > 0 then2
a2(2a+b)2+ 2
b2(2b+a)2+ 1
(a2+2b2)2+ 1
(b2+2a2)2≥ (2a+b)2
9(2a6+b6)+ (2b+a)2
9(2b6+a6).
Mihaly Bencze
PP29522. If a, b, c > 0 and abc ≤ 1 then�2a + a
bc
�2+�2b +
bca
�2+�2c +
cab
�2 ≥ 27.
Mihaly Bencze
PP29523. If a, b, c > 0 ad λ > 0 thenP 1
(λa2+bc)2≥ 27
(λ+1)2(�
a2)2.
Mihaly Bencze
PP29524. In all triangle ABC holdsra
√s
rb√ra+4s
+ rrb√s
rc√rb+4s
+ src√r
rra√ra+4r
≥q
35
�4p
ras + 4
prbs + 4
prcs
�and his
permutations.
Mihaly Bencze
1006 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP29525. In all triangle ABC holds
Rrha
hb
√2(Rha+8s)
+ hb
√r
hc
√s(hb+4sr)
+ hc
Rha
√r(hc+4r)
≥s
35
�4
qRha
2s + 4
qhb
sr + 4
qhc
r
�
and his permutations.
Mihaly Bencze
PP29526. If a, b, c > 0 then�P a4c4
b6√a4+4b4
��P a8
b4c2√a4+4c4
�≥ 3
5abc
p(P
a2b) (P
a2c).
Mihaly Bencze
PP29527. In all triangle ABC holds
aR
2b√
s(a+16s)+ br
c√
R(b+4R)+ 4cs
a√
r(c+4r)≥s
35
�4p
a4s +
4
qbR + 4
pcr
�and his
permutations.
Mihaly Bencze
PP29528. If a, b > 0 thena4(1+a4b8)b4√a4+4
+b4(1+b4a8)a4
√b4+4
+ 2a2b2
√1+4a4b4
≥ 2
q3((a+b)ab+1)
5 .
Mihaly Bencze
PP29529. If ak > 0 (k = 1, 2, ..., n) thenPcyclic
a21(a2+a3+...+an)
2√
a41+4≥ n2
√2m+1
(n−1)2�
(m+1)n�
k=1a2k−2m
n�
k=1ak+2n(m+1)
� for all
n,m ∈ N,n ≥ 2.
Mihaly Bencze
PP29530. Determine all a, b > 0 for which
ab +
b
4(√a+
√b)
2 +4(
√a+
√b)
2
a ≥ ab+ 4 (a+ b)�√
a+√b�2
.
Mihaly Bencze
PP29531. If a, b, c > 0 andP b
3a+1 ≤ 8, thenP ab
80a3+16a2+1≤ 1.
Mihaly Bencze
Proposed Problems 1007
PP29532. In all triangle ABC holds
1).P�
rbra
�2≥ 81r2((4R+r)2−2s2)
s4
2).P�
hb
ha
�2≥ 81r4
��s2+r2+4Rr
4s2r2
�2− R
s2r3
�
Mihaly Bencze
PP29533. If x ∈�0, π2
�then tg4x+ cos4 x+ 1
sin4 x≥
�1− sin2 x cos2 x
�2.
Mihaly Bencze
PP29534. If a, b, c > 0 and abc=1 thenP 3a4+2b2−4a+11
(ab+a+1)2≥ 4.
Mihaly Bencze
PP29535. If a, b, c > 0 and abc = 1 thenP 1
(3a4+2b2−4a+11)n≥ 3
12n for all
n ∈ N.
Mihaly Bencze
PP29536. If a, b > 0 then1
3√3a4+2b2−4a+11+ 3
qa2b2
3a2b6+11a2b2−4a2b3+2+ 3
qa4b4
11a4b4+2a6b4−4a3b3+3≤ 3
q94 .
Mihaly Bencze
PP29537. If ak > 0 (k = 1, 2, ..., n) , thenQ
cyclic
3a41+2a22−4a1+11a1a2+a1+1 ≥ 4n.
Mihaly Bencze
PP29538. If x ∈�0, π2
�then
(1 + sinx+ cosx)2 + sin8 x+ cos8 x ≥ 53 + 2
3 sin2 2x.
Mihaly Bencze
PP29539. Determine all x, y > 0 for which
3
�x4 + y4 + 16 (x+ y)4 +
�√x+
√y +
p2 (x+ y)
�2�
≥
≥ 4�5x2 + 8xy + 5y2
� �2x2 + 5xy + 2y2
�.
Mihaly Bencze
1008 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP29540. If xk > 0 (k = 1, 2, ..., n) andnP
k=1
xk = n then
nPk=1
x4k +
�nP
k=1
√xk
�2
≥ n+1n
�nP
k=1
x2k
� Pcyclic
x1x2
!.
Mihaly Bencze
PP29541. In all triangle ABC holds:
1). 27r3P 1
r4a+�P 1√
ra
�2≥ 36
s2
�s2 − 2r (4R+ r)
�(4R+ r)
2). 27r3P 1
h4a+�P 1√
ha
�2≥ 9(s2−r2−4Rr)(s2+r2+4Rr)
2s4r
Mihaly Bencze
PP29542. If ak > 0 (k = 1, 2, ..., n) andnQ
k=1
ak = 1 then
Pcyclic
am1
am+12
�
r(am1 +r−1)≥ n
r for all m ∈ N∗, r > 1.
Mihaly Bencze
PP29543. If a, b > 0 then a4
b4√
5(a4+4)+ a5b9√
5(b4+4)+ 1
a7b2√5+4a4b4
≥ 35 .
Mihaly Bencze
PP29544. If x ∈ R then�2 + sin2 x
�p2 + sin2 x+
�2 + cos2 x
�√2 + cos2 x ≥ 3
√2− 1.
Mihaly Bencze
PP29545. Determine all a, b > 0 for whicha√a+ b+ b
√2a+ 3b+ 2 (a+ b)
√3a+ 2b ≥ 3
√2.
Mihaly Bencze
PP29546. In all triangle ABC holds
1).P 1
ra
q1ra
+ 1rb
≥√2
r√3r
2).P 1
ha
q1ha
+ 1hb
≥√2
r√3r
Mihaly Bencze
Proposed Problems 1009
PP29547. If x, y > 0 then53xy (x+ y) + x4y3
x3y6+x3y3+1+ x3y4
x6y3+x3y3+1+ 1
xy(x3+y3+1)≥ 6.
Mihaly Bencze
PP29548. If ak > 0 (k = 1, 2, ..., n) andnP
k=1
ak = n thenP
cyclic
a1√a1 + a2 + ...+ an−1 ≥ n
√n− 1.
Mihaly Bencze
PP29549. Determine all a, b > 0 for which12ab+ 60 (a+ b)2 + 5ab (a+ b) ≥ a3 + b3 + 125 (a+ b)3 + 102.
Mihaly Bencze
PP29550. If xk > 0 (k = 1, 2, ..., n) and λ ≥ 1 andnQ
k=1
xk = 1 then
λnP
k=1
xk +P
cyclic
x1xn2+...+xn
n+1 ≥ λ+ 1.
Mihaly Bencze
PP29551. If ak > 0 (k = 1, 2, ..., n) then
Pcyclic
a1a2...an−1
n−1�
(λ+an−11 )(λ+an−1
2 )...(λ+an−1n−1)
≤n
�
cyclic
a1a2...an−1
λn+�
cyclic
a1a2...an−1for all λ > 0.
Mihaly Bencze
PP29552. Determine all x ∈ R for which17 sin2 x cos2 x ≥ sin6 x+ cos6 x+ 16.
Mihaly Bencze
PP29553. If ak > 0 (k = 1, 2, ..., n) and ai 6= aj(i = 1, 2, ..., n; j = 1, 2, ..., n, i 6= j) thenP
1≤i<j≤n
ai+aj(ai−aj)
2 + (n− 1)nP
k=1
1ak
≥ 9(n−1)n2
4n�
k=1ak
.
Mihaly Bencze
1010 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP29554. If ak > 0 (k = 1, 2, ..., n) and ai 6= aj
(i = 1, 2, ..., n; j = 1, 2, ..., n, i 6= j) thenP
1≤i<j≤n
�1
(ai−aj)2 + 1
aiaj
�≥ 9(n−1)n2
8n�
k=1a2k
.
Mihaly Bencze
PP29555. If a, b, c > 0 and a 6= b 6= c thenP a+2b+3c
(a+2b−3c)2+ 9
P 1a + 3
P abc ≥ 3(37+4
√3+4
√6+24
√2)
2�
a .
Mihaly Bencze
PP29556. If ak > 0 (k = 1, 2, ..., n) and ai 6= aj(i = 1, 2, ..., n; j = 1, 2, ..., n, i 6= j) thenP1≤i<j≤n
�1
(ai−aj)2 + 1
aiaj
�≥ 9n2
4
�
n�
k=1ak
�2 .
Mihaly Bencze
PP29557. If a, b, c > 0 and a 6= b 6= c thenP 1(a+2b−3c)2
+ 2P 1
ab ≥�37 + 4
√3 + 4
√6 + 24
√2�P 1
(a+2b+3c)2.
Mihaly Bencze
PP29558. If a, b, c > 0 and a 6= b 6= c thenP a+b+2c
(a+b−2c)2+ 4
P 1a + 2
�
a2
abc ≥ 9(33+8√2)
4�
a .
Mihaly Bencze
PP29559. If ak > 0 (k = 1, 2, ..., n) and m ∈ N∗ thenPcyclic
a1+a2
(am1 3√a2+am2
3√a1)
3 ≥ n2
4n�
k=1amk
Mihaly Bencze
PP29560. If a, b, c > 0 thenP 1
(a+b−2c)2+ 2
P 1ab ≥
�33 + 8
√2�P 1
(a+b+2c)2
for all a 6= b 6= c.
Mihaly Bencze
Proposed Problems 1011
PP29561. If xk ∈�0, π2
�(k = 1, 2, ..., n) then
Pcyclic
�1+sinx1 cosx2sinx1+cosx2
�2≤ 9n
8 .
Mihaly Bencze
PP29562. If a, b, c > 0 thenP a+b
(an 3√b+bn 3√a)
3 ≥ 94�
a3nfor all n ∈ N.
Mihaly Bencze
PP29563. In all triangle ABC holds
1).Q�
a+√ab+b√
a+√b
�2≤ 729s(s2+r2+2Rr)
256
2).Q�
c+√
(s−a)(s−b)√s−a+
√s−b
�2
≤ 729sRr128
3).Q�
ra+√rarb+rb√
ra+√rb
�2≤ 729s2R
128
4).Q�
ha+√hahb+hb√
ha+√hb
�2≤ 729s2r(s2+r2+2Rr)
512R2
5).Q�
sin2 A2+sin A
2sin B
2+sin2 B
2
sin A2+sin B
2
�2
≤ 729((2R−r)(s2+r2−8Rr)−2Rr2)16384R3
6).Q�
cos2 A2+cos A
2cos B
2+cos2 B
2
cos A2+cos B
2
�2
≤ 729((4R+r)3+s2(2R+r))16384R3
Mihaly Bencze
PP29564. In all triangle ABC holds
1).P
b�a+
√ac+c√
a+√c
�2≤ 9
4
p(s2 − r2 − 4Rr) (3s2 − r2 − 4Rr)
2).P
(s− b)
�b+√
(s−a)(s−c)√s−a+
√s−c
�2
≤ 94√2
p(s2 − 2r2 − 8Rr) (s2 − r2 − 4Rr)
3).P
rb
�ra+
√rarc+rc√
ra+√rc
�2≤ 9
4√2
r�(4R+ r)2 − 2s2
��(4R+ r)2 − s2
�
4).P
hb
�ha+
√hahc+hc√
ha+√hc
�2≤
≤ 9
4√2
vuut �
s2 + r2 + 4Rr
2R
�2
− 4s2r
R
! �s2 + r2 + 4Rr
2R
�2
− 2s2r
R
!
Mihaly Bencze
1012 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP29565. In all triangle ABC holds
1).P 1
b+c
� √a+
√b
a+√ab+b
�2≥ 16
9(s2+r2+2Rr)
2).P 1
a
� √s−a+
√s−b
c+√
(s−a)(s−b)
�2
≥ 49Rr
3).P 1
ra+rb
� √ra+
√rb
ra+√rarb+rb
�2≥ 4(4R+r)
9s2R
4).P 1
ha+hb
� √ha+
√hb
ha+√hahb+hb
�2≥ 8R(s2+r2+4Rr)
9s2r(s2+r2+2Rr)
5).P 1
sin2 B2+sin2 C
2
�sin A
2+sin B
2
sin2 A2+sin A
2sin B
2+sin2 B
2
�2
≥ 256R2(2R−r)9((2R−r)(s2+r2−8Rr)−2Rr2)
6).P 1
cos2 B2+cos2 C
2
�cos A
2+cos B
2
sin2 A2+sin A
2sin B
2+sin2 B
2
�2
≥ 256R2(4R+r)
9((4R+r)3+s2(2R+r))
Mihaly Bencze
PP29566. If x ∈�0, π2
�then 4
3 + 1+ 3√2+3√sin2 x+
3√cos2 x
4−√2−(sinx+cosx)
≥ 163 sin2 x cos2 x.
Mihaly Bencze
PP29567. Determine all a, b > 0 for which3√a+
3√b+ 3
√2(a+b)
4−�√
a+√b+√
2(a+b)� + 8
3
�5a2 + 8ab+ 5b2
�≥ 12.
Mihaly Bencze
PP29568. In all triangle ABC holds
1).1+ 3√3r
� 13√ra
4−√3r
� 1√ra
+ 12 ≥ 48r(4R+r)s2
2).1+ 3√3r
� 13√ha
4−√3r
� 1√ha
≥ 12r(4R+r)s2
Mihaly Bencze
PP29569. Prove that1R
−1
�
n�
k=1
xk2+k−1
�
exdx
1+ex = nn+1 .
Mihaly Bencze
PP29570. In all triangle ABC holds
1).P a+4
√ab+b
(a−√ab+b)
2 ≤ 6(5s2+r2+4Rr)s(s2+r2+2Rr)
Proposed Problems 1013
2).P c+4
√(s−a)(s−b)
�
c−√
(s−a)(s−b)�2 ≤ 3(s2+r2+4Rr)
sRr
3).P ra+4
√rarb+rb
(ra−√rarb+rb)
2 ≤ 3((4R+r)2+s2)s2R
4).P ha+4
√hahb+hb
(ha−√hahb+hb)
2 ≤3�
(s2+r2+4Rr)2+8s2Rr
�
s2r(s2+r2+2Rr)
5).P sin2 A
2+4 sin A
2sin B
2+sin2 B
2
(sin2 A2−sin A
2sin B
2+sin2 B
2 )2 ≤ 24R(16R2−24Rr+5r2+s2)
(2R−r)(s2+r2−8Rr)−2Rr2
6).P cos2 A
2+4 cos A
2cos B
2+cos2 B
2
(cos2 A2−cos A
2cos B
2+cos2 B
2 )2 ≤ 24R(5(4R+r)2+s2)
(4R+r)3+s2(2R+r)
Mihaly Bencze
PP29571. In all triangle ABC holds
1).P cos3 A
2
cos B2+cos C
2−cos A
2
≥ 4R+r2R
2).P cos5 A
2
cos B2+cos C
2−cos A
2
≥ (4R+r)2−s2
8R2
3).P m3
a
mb+mc−ma≥ 3
2
�s2 − r2 − 4Rr
�
4).P m5
a
mb+mc−ma≥ 9
8
�2�s2 − r2 − 4Rr
�2 −�s2 + r2 + 4Rr
�2+ 16s2Rr
�
Mihaly Bencze
PP29572. In all triangle ABC holds
1).P a2+ab+b2
a3+b3≤ 3(5s2+r2+4Rr)
2s(s2+r2+2Rr)
2).P (s−a)2+(s−a)(s−b)+(s−b)2
(s−a)3+(s−b)3≤ 3(s2+r2+4Rr)
4sRr
3).P r2a+rarb+r2
b
r3a+r3b
≤ 4((4R+r)2+s2)4s2R
4).P h2
a+hahb+h2b
h3a+h3
b
≤3�
(s2+r2+4Rr)2+8s2Rr
�
4s2r(s2+r2+2Rr)
5).P sin4 A
2+sin2 A
2sin2 B
2+sin4 B
2
sin6 A2+sin6 B
2
≤ 6R(16R2−24Rr+5r2+s2)(2R−r)(s2+r2−8Rr)−2Rr2
6).P cos4 A
2+cos2 A
2cos2 B
2+cos4 B
2
cos6 A2+cos6 B
2
≤ 6R(5(4R+r)2+s2)(4R+r)3+s2(2R+r)
Mihaly Bencze
PP29573. If ak > 0 (k = 1, 2, ..., n) thenP
1≤i<j≤n
�a3iaj
+a3jai
�≥ (n− 1)
nPk=1
a2k.
Mihaly Bencze
1014 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP29574. If ak > 0 (k = 1, 2, ..., n) then
1).P
cyclic
(5a1+4a2)(5a1+a2)a1+a2
≥ 27nP
k=1
ak
2).P
cyclic
(5a1+4a2)(5a1+a2)8a1+a2
≥ 6nP
k=1
ak
Mihaly Bencze
PP29575. In all triangle ABC holds
1).P sinA
(1−sinA)2≥
√6�10− s2+r2+4Rr
2sr
�
2).P sin2 A
cos4 A≥
√6
�10 + 4R
r −�s2+r2+Rr
2sr
�2�
Mihaly Bencze
PP29576. In all acute triangle ABC holds
1).P cosA
(1−cosA)2≥
√6�10− s2+r2−4R2
s2−(2R+r)2
�
2).P cos2 A
sin4 A≥
√6
�10 + 8R(R+r)
s2−(2R+r)2−�
s2+r2−4R2
s2−(2R+r)2
�2�
Mihaly Bencze
PP29577. If ak > 0 (k = 1, 2, ..., n) andnP
k=1
ak = 1 then
P a51(a2+1)(a3+1) ≥
1n2(n+1)2
.
Mihaly Bencze
PP29578. In all triangle ABC holds1).
P rbrc(r+rb)(r+rc)r5a
≥ 1144r5
2).P hbhc
(r+hb)(r+hc)h5a≥ 1
144r5
Mihaly Bencze
PP29579. In all triangle ABC holds1).
P Arb
≥ π3r
2).P A
hb≥ π
3r
Mihaly Bencze
Proposed Problems 1015
PP29580. If xk ∈�0, π2
�(k = 1, 2, ..., n) then
Pcyclic
(1−sinx1 cosx2)2
1+4 sinx1 cosx2≥ n
12 .
Mihaly Bencze
PP29581. Prove thatnQ
k=1
((k + 1)!)k2 ≤
�6((n−1)(n+2)!+2)n(n+1)(2n+1)
�n(n+1)(2n+1)6
.
Mihaly Bencze
PP29582. Prove thatnP
k=1
k! ≤ 12((n−1)(n+2)!+2)(n+1)(n+3)(2n+1) .
Mihaly Bencze
PP29583. In all triangle ABC holds1).
P 2a2−3ab+2b2
a4+b4≤ 1
4Rr
2).P 2(s−a)2−3(s−a)(s−b)+2(s−b)2
(s−a)4+(s−b)4≤ 1
2r2
3).P 2r2a−3rarb+2r2
b
r4a+r4b
≤ 4R+r4s2r
4).P 2h2
a−3hahb+2h2b
h4a+h4
b
≤ s2+r2+4Rr8s2r2
5).P 2 sin4 A
2−3 sin2 A
2sin2 B
2+sin4 B
2
sin8 A2+sin8 B
2
≤ 4R(2R−r)r2
6).P 2 cos4 A
2−3 cos2 A
2cos2 B
2+cos4 B
2
cos8 A2+cos8 B
2
≤ 4R(4R+r)s2
Mihaly Bencze
PP29584. In all triangle ABC holds
6 +P �
tg3A2 ctgB2 + tg3B2 ctg
A2
�≥ 8(4s2r2−2(s2−r2−4Rr)(s2−r2−4Rr−4R2))
(s2−(2R+r)2)2 .
Mihaly Bencze
PP29585. In all triangle ABC holdsP (2a2−3ab+2b2)(a+b)
a4+b4≤ s2+r2+4Rr
4sRr .
Mihaly Bencze
PP29586. Prove that (n+1)(n+2)3 ≤ (n−1)(n+2)!+2
(n+1)!−1 .
Mihaly Bencze
1016 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP29587. Prove thatnP
k=1
k! ≤ 2((n+1)!−1)n+1 .
Mihaly Bencze
PP29588. Prove thatnP
k=1
1(k+1)! ≤
2((n+1)!−1)(n+1)(n+1)! .
Mihaly Bencze
PP29589. Prove thatnP
k=1
k! ≤ 3(n+1)(n+1)!n2+3n+5
.
Mihaly Bencze
PP29590. Prove thatnP
k=1
14k4+1
≤ n2n2+2n+1
.
Mihaly Bencze
PP29591. In all triangle ABC holdsP (2a2−3ab+2b2)(a+b)2
a4+b4≤ s2+r2+10Rr
4Rr .
Mihaly Bencze
PP29592. In all triangle ABC holdsP a4+b4
(a+b)(2a2−3ab+2b2)≥ (s2+r2+4Rr)
2+8s2Rr
s(s2+r2+2Rr).
Mihaly Bencze
PP29593. In all triangle ABC holdsP a4+b4
(2a2−3ab+2b2)(b+c)(c+a)≥ 2(s2+r2−2Rr)
s2+r2+2Rr.
Mihaly Bencze
PP29594. If ak > 0 (k = 1, 2, ..., n) , thenP
1≤i<j≤n
a4i+a4jaiaj
≥ (n− 1)nP
k=1
a2k.
Mihaly Bencze
PP29595. If ak > 0 (k = 1, 2, ..., n) , thenP
1≤i<j≤naiaj
�2a2i − 3aiaj + 2a2j
�≤ n−1
2
nPk=1
a2k.
Mihaly Bencze
Proposed Problems 1017
PP29596. In all scalene triangle ABC holds
1).P
w2a
qm2
a−h2a
w2a−h2
a= s2 + r2 + 4Rr
2).P w2
ahb
ha
qm2
a−h2a
w2a−h2
a= s2 + r2 + 4Rr
3).P
w2aha
qm2
a−h2a
w2a−h2
a=
(s2+r2+4Rr)2−16s2Rr
2R
4).P w2
araha
qm2
a−h2a
w2a−h2
a= 2R (4R+ r)
5).P w2
ar2a
ha
qm2
a−h2a
w2a−h2
a= 2R
�(4R− r)2 − 2s2
�
Mihaly Bencze
PP29597. If a, b ∈ Z such that a2 + b2 is divisible by a+ b2, then2a3b2 + ab2 + 3b4 is divisible by a+ b2.
Mihaly Bencze
PP29598. In all triangle ABC holds
2P
ab sin A2 sin B
2 ≥�s+ s(R−2r)2
4R(R+2r)
�2− (4R+r)r2+(2R−3r)s2
2R .
Mihaly Bencze
PP29599. In all acute triangle ABC holdsP ln(9−cosA)−ln(cosA)
ln(9−cosA)−ln(3−cosA) ≤3(s2+r2−4R2)2(s2−(2R+r)2)
.
Mihaly Bencze
PP29600. In all triangle ABC holds
1).P ln(8+cos2 A
2 )−2 ln(sin A2 )
ln(8+cos2 A2 )−ln(2+cos2 A
2 )≤ 3(s2+r2−8Rr)
2r2
2).P ln(8+sin2 A
2 )−2 ln(cos A2 )
ln(8+sin2 A2 )−ln(2+sin2 A
2 )≤ 3(s2+(4R+r)2)
2s2
Mihaly Bencze
PP29601. If x, y ∈ [0, 1] then�3− x2
� �3− y2
�sin (x+ y) ≥ 3 (x+ y) (3− xy) cosx cos y.
Mihaly Bencze
PP29602. In all triangle ABC holdsP ln(9−sinA)−ln(sinA)
ln(9−sinA)−ln(3−sinA) ≤3(s2+r2+4Rr)
4sr .
Mihaly Bencze
1018 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP29603. In all acute triangle ABC holds 9− π2
3 ≥ 3(R+r)R + 2
PA sinA.
Mihaly Bencze
PP29604. If x ∈�0, π2
�then sinx
tg(sinx) +cosx
tg(cosx) ≤53 .
Mihaly Bencze
PP29605. In all acute triangle ABC holdsP �2 + sin2A
�tg (cosA) ≥ 3(R+r)
R .
Mihaly Bencze
PP29606. In all triangle ABC holds1).
P �3− sin4 A
2
�tg
�sin2 A
2
�≥ 3(2R−r)
2R
2).P �
3− cos4 A2
�tg
�cos2 A
2
�≥ 3(4R+r)
2R
Mihaly Bencze
PP29607. If xk ∈ [0, 1] (k = 1, 2, ..., n) thennP
k=1
tgxk ≥3n2
n�
k=1xk
3n2−�
n�
k=1xk
�2 .
Mihaly Bencze
PP29608. If x ∈�0, π2
�then�
3− sin4 x�tg
�sin2 x
�+�3− cos4 x
�tg
�cos2 x
�≥ 3.
Mihaly Bencze
PP29609. If a, b > 0 then
�2(a+b)2
ab
�2a2+b2
+�2(a+b)2
ab
�a2+2b2
≤�3a2+2ab+b2
a(a+2b)
�(2a+b)2
+�a2+2ab+3b2
b(b+2a)
�(a+2b)2
.
Mihaly Bencze
PP29610. If ak ∈ [0, 1] (k = 1, 2, ..., n) thennP
k=1
aλk�3− a2k
�tgak ≥ 3
nλ
�nP
k=1
ak
�λ+1
for all λ ∈ (−∞,−1] ∪ [0,+∞) .
Mihaly Bencze
Proposed Problems 1019
PP29611. In all triangle ABC holdsP
Aλ−1�3π2 −A2
�tgA
π ≥ πλ+1
3λ−2 for allλ ∈ (−∞,−1] ∪ [1,+∞) .
Mihaly Bencze
PP29612. In all triangle ABC holdsP �
2 + cos2A�tg (sinA) ≥ 3s
R .
Mihaly Bencze
PP29613. Prove thatnP
k=1
�9−
�kn
�4�tg k
n ≥ 3(n+1)(7n+1)4n .
Mihaly Bencze
PP29614. Prove thatnP
k=1
kctg kn ≤ 16n2−3n−1
18 .
Mihaly Bencze
PP29615. If 0 ≤ a ≤ b ≤ 1 then�cos acos b
� 23 ≥ 3−a2
3−b2.
Mihaly Bencze
PP29616. If a, b > 0 then 9�4a3 + b (a+ b)3
��4b3 + a (a+ b)3
�≥
≥ (2a+ b)2 (a+ 2b)2 (a+ b)4 .
Mihaly Bencze
PP29617. If a, b, c > 0, λ > 0 such thatP
a = 9λ+1 thenp
λP
a2 + (λ2 + λ+ 1)P
ab ≤P√a+ λb.
Mihaly Bencze
PP29618. If xk ∈ R (k = 1, 2, ..., n) , xi 6= xj(i 6= j) andnP
k=1
xk = 1 then
Pcyclic
ex1−ex2x1−x2
≥ ne1n .
Mihaly Bencze
PP29619. Let ABCD be a tetrahedron inscribed in a sphere with radius R.The insphere of the tetrahedron have the incenter I and inradius r. DenoteRA, RB, RC , RD the radii of the circumsphere of tetrahedronsIBCD, IACD, IABD, IABC.
1020 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
Prove that V ol[RARBRCRD]V ol[ABCD] = R
3r .
Mihaly Bencze
PP29620. If a, b, c > 0 and λ > 0 such thatP
a = 3λ+1 thenP 1
1+2(a+λb)(b+λc) ≥2
1+(a+λb)(b+λc)(c+λa) .
Mihaly Bencze
PP29621. If x ∈�0, π2
�then 1+ 4
1+sin 2x+4 cos2 x+ 4
1+sin 2x+4 sin2 x≥ 40
9(1+sin 2x) .
Mihaly Bencze
PP29622. Let A1A2...An be a convex polygon, and M a random point inspace. Denote x = A1MA2∡ = A2MA3∡ = ... = AnMA1∡. DenoteR1, R2, ..., Rn the circumradii of triangles A1MA2, A2MA3, ..., AnMA1.
Determine x such that
�nP
k=1
Rk
�2
≥nP
k=1
MA2k.
Mihaly Bencze
PP29623. If a, b > 0 and (a+ b)√a = 2, then (2a+ b)3 + 5a2b ≥ 32.
Mihaly Bencze
PP29624. If a, b, c > 0 and a3 + ab (a+ b)− 2a− b = 0 then2a2
�2 + a2
�≥ (2 + ab)
�2 + a2 − b2
�.
Mihaly Bencze
PP29625. In all triangle ABC holdsP (a+b)2
c ≤ 6√3R2.
Mihaly Bencze
PP29626. In all triangle ABC holds1).
�r3a − r3
� �r3b − r3
� �r3c − r3
�≥ 18252r9
2).�h3a − r3
� �h3b − r3
� �h3c − r3
�≥ 18252r9
Mihaly Bencze
PP29627. In all triangle ABC holdsP �
3π2 −A2�tgA
π ≥ 3π2.
Mihaly Bencze
Proposed Problems 1021
PP29628. If a, b > 0 and n ≥ 3 then�(2a+ b)n−1 + n− 1
��(a+ 2b)n−1 + n− 1
�≥ 9n2ab
n√2a2+5ab+2b2
, for all n ∈ N∗.
Mihaly Bencze
PP29629. If x, y, z > 0 then
27(ex−1)(ey−1)(ez−1)xyz
�ex+y+z
3 −1x+y+z
�3
≥ 64
�
ex+y2 −1
��
ey+z2 −1
��
ez+x2 −1
�
(x+y)(y+z)(z+x)
2
.
Mihaly Bencze
PP29630. If x ∈�0, π2
�then 16 + 7
sin4 x cos4 x≤ 4
sin2 x cos2 x.
Mihaly Bencze
PP29631. If a, b, c > 0 then 6P
(a− bc)2 ≥ (P |a− b|)2 .
Mihaly Bencze
PP29632. Find all ak, bk ∈ R (k = 1, 2, ..., n) such that
3x5+4x4+5x3+6x2+7x+8(x2−x+1)n
=nP
k=1
akx+bk(x2−x+1)k
.
Mihaly Bencze
PP29633. If x > 0 then ch6x+ sh6x+ 1ch2x
+ 1sh2x
+ ch3xshx + sh3x
chx ≥ 32 .
Mihaly Bencze
PP29634. If x, y, z, t > 0 then
2x+y+z+t + 2x+y + 2x+z + 2x+t + 2y+z + 2y+t + 2z+t + 1 ≥≥ 2x + 2y + 2z + 2t + xy+yz+zx
√512xyz.
Mihaly Bencze
PP29635. If ak ∈ (−λ,λ) ,λ > 0 (k = 1, 2, ..., n) , thenP |a1+a2|+|an+a1|λ−a21
≥ 8P |a+a2|
4λ−(a1+a2)2 .
Mihaly Bencze
PP29636. In all triangle ABC holds1).
P 1cos2 A
≥ 4P 1
4−(sinA+sinB)2
2).P 1
sin2 A≥ 4
P 14−(cosA+cosB)2
Mihaly Bencze
1022 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP29637. If ai ∈ (−λ,λ) , (i = 1, 2, ..., n) , λ > 0 and k ∈ {1, 2, ..., n} , thennP
i=1
1λ−a2i
≥ k2P 1
λk2−(a1+...+ak)2 .
Mihaly Bencze
PP29638. In all convex quadrilateral ABCD holds
1).P 1
cos2 A≥ 9
P 19−(sinA+sinB+sinC)2
2).P 1
sin2 A≥ 9
P 19−(cosA+cosB+cosC)2
Mihaly Bencze
PP29639. In all convex polygon A1A2...An holds
1).nP
i=1
1cos2 Ai
≥ k2P 1
k2−(sinA1+...+sinAk)2
2).nP
i=1
1sin2 Ai
≥ k2P 1
k2−(cosA1+...+cosAk)2
Mihaly Bencze
PP29640. If xk ∈ (0, 1) (k = 1, 2, ..., n) and λ ∈ (−∞, 0] ∪ [1,+∞) and
k ∈ {1, 2, ..., n} , thennP
i=1
11−xλ
i
≥ kλP 1
kλ−(x1+...+xk)λ .
Mihaly Bencze
PP29641. If x, y, ak > 0 (k = 1, 2, ..., n) then
n
snQ
k=1
(xak + y) ≥ y +nx
n�
k=1ak
�
cyclic
a1a2...an−1.
Mihaly Bencze
PP29642. If x > 0 then144 (x cosx− sinx)2 (x sinx+ cosx− 1)2 ≤ x6
�4x2 − sin2 2x
�.
Mihaly Bencze
PP29643. If x > 0 then 3 (x sinx+ cosx− 1)2 + 3 (sinx− x cosx)2 ≤ x4.
Mihaly Bencze
Proposed Problems 1023
PP29644. If ai > 0 (i = 1, 2, ..., n) and k, p ∈ N∗ thennQ
i=1
�a2(k+p)i − a2ki + 2p+ 1
�≥
�nP
i=1
2p+1
qa2pi
�2p+1
.
Mihaly Bencze
PP29645. Let ABCDE be a convex pentagon. Prove that:
√AB2 +BC2 + CD2 +
√BC2 + CD2 +DA2 +
√CD2 +DA2 +AB2+
+√DA2 +AB2 +BC2 +
√BC2 + CD2 +DE2 +
√CD2 +DE2 + EB2+
+√DE2 + EB2 +BC2 +
√EB2 +BC2 + CD2 +
√CD2 +DE2 + EA2+
+√DE2 + EA2 +AC2 +
√EA2 +AC2 + CD2 +
√AC2 + CD2 +DE2+
+√DE2 + EA2 + EB2 +
√EA2 + EB2 +BD2 +
√EB2 +BD2 +DE2+
+√BD2 +DE2 + EA2 +
√EA2 +AB2 +BC2 +
√AB2 +BC2 +DE2+
+√BC2 +DE2 + EA2 +
√DE2 + EA2 +AB2 ≥
≥ 2√3�√
AC ·BD +√BD · CE +
√CE ·DA+
√DA · EB +
√EB ·AC
�.
Mihaly Bencze
PP29646. If ak > 0 (k = 1, 2, ..., n) then
P a1a2+a3+...+an
≥
�
n�
k=1ak
�2
2�
1≤i<j≤n
aiaj≥ n
n−1 .
Mihaly Bencze
PP29647. Let ABCD be a cyclic quadrilateral. Prove thatsinA sinB + sinB sinC + sinC sinD + sinD sinA ≤≤ 4
abcd (−a+ b+ c+ d) (a− b+ c+ d) (a+ b− c+ d) (a+ b+ c− d) .
Mihaly Bencze
PP29648. Let ABCDE be a convex pentagon inscribed in circle with radiusR. Prove that3 (AB +BC + CD +DE + EA) +DA+ EB +AC +BD + CE ≥≥ 2
q2√2
R ((AB ·BC · CD ·DA)38 + (BC · CD ·DE · ED)
38 +
+(CD ·DE · EA ·AC)38 + (DE · EA ·AB ·BD)
38 + (EA ·AB ·BC · CE)
38 ).
Mihaly Bencze
1024 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP29649. Determine all cyclic quadrilaterals ABCD in which
cos A2 + cos B
2 + cos C2 + cos D
2 ≤ 44
qArea2[ABCD]
4abcd .
Mihaly Bencze
PP29650. In the cyclic quadrilateral ABCD holds
�a2 − b2 − c2 + d2 + F
� �b2 − c2 − d2 + a2 + F
� �c2 − d2 − a2 + b2 + F
�·
·�d2 − a2 − b2 + c2 + F
�≤ 64 (ad+ bc)4 where F = Area [ABCD] .
Mihaly Bencze
PP29651. If ak > 0 (k = 1, 2, ..., n) , thenP
cyclic
a21−a1a2+a22a22+a2a3+a23
≥ n3 .
Mihaly Bencze
PP29652. If ak > 0 (k = 1, 2, ..., n) , thenP (a21−a1a2+a22)
2
(a22+a2a3+a23)(a23+a3a4+a24)≥ n
9 .
Mihaly Bencze
PP29653. Prove thatnP
k=1
3k2+3k+1k(k3+2k2+2k+1)
≤ 3nn+1 .
Mihaly Bencze
PP29654. If a, b, c > 0 then 9 ≥ (2�
a2+�
ab)2
2�
a4−�
a2b2.
Mihaly Bencze
PP29655. If ak > 0 (k = 1, 2, ..., n) thenP a41+a21a
22+a42
(a22+a2a3+a23)(a23+a3a4+a24)≥ n
3 .
Mihaly Bencze
PP29656. In all triangle ABC holds
1).P m2
a−mamb+m2b
m2b+mbmc+m2
c< 3
2).P 2+cosA+cosB−2 cos A
2cos B
2
2+cosB+cosC−2 cos B2cos C
2
< 3
Mihaly Bencze
Proposed Problems 1025
PP29657. If ak > 0 (k = 1, 2, ..., n) , thenP�
a21−a1a2+a22a22+a2a3+a23
�λ≥ n
3λfor all
λ ∈ (−∞, 0] ∪ [1,+∞) .
Mihaly Bencze
PP29658. In all triangle ABC holds
1).Q
(ma +mb) ≥ 89 (P
ma) (P
mamb)2). 9
Qcos
�π−A4
�Qcos A−B
4 ≥�P
cos A2
� �Pcos A
2 cos B2
�
Mihaly Bencze
PP29659. In all acute triangle ABC holds
1).P 1
cosA(5+4 cosA) ≥67 − 18
49 ln
�2(s2−(2R+r)2)
R2
�
2).P cos2 A
1+5 cosA ≥ 314 + 9
98 ln
�2(s2−(2R+r)2)
R2
�
Mihaly Bencze
PP29660. In all triangle ABC holds1).
P 1ra(5+2ra)
≥ 37 − 9
49 ln�s2r
�
2).P r2a
2+5ra≥ 3
7 + 949 ln
�s2r
�
Mihaly Bencze
PP29661. In all triangle ABC holds
1).P 1
sinA(5√3+4 sinA)
≥ 221 − 2
49 ln�
4sr3√3R2
�
2).P sin2 A
5 sinA+√3≥ 2
√3
7 + 6√3
49 ln�
4sr3√3R2
�
Mihaly Bencze
PP29662. If ak > 0 (k = 1, 2, ..., n) , thennP
k=1
1xk(5+2xk)
≥ n7 − 9
49 ln
�nQ
k=1
ak
�.
Mihaly Bencze
1026 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP29663. Prove that
1).nP
k=1
(k+1)2
k(7k+5) ≥n7 + 9
49 ln (n+ 1)
2).nP
k=1
k2
(k+1)(7k+2) ≥n7 − 9
49 ln (n+ 1)
Mihaly Bencze
PP29664. If ak > 0 (k = 1, 2, ..., n) , thenP a22
a1(2a1+5a2)≥ n
7 .
Mihaly Bencze
PP29665. Prove that
1).nP
k=1
k2k
(k+1)k(5kk+2(k+1)k)≥ n
7 − 949 ln
�(n+1)n
n!
�
2).nP
k=1
(k+1)2k
kk(2kk+5(k+1)k)≥ n
7 + 949 ln
�(n+1)n
n!
�
Mihaly Bencze
PP29666. If λ > 0 thennP
k=1
1kλ(5+2kλ)
≥ n7 − 9λ
49 ln (n!) .
Mihaly Bencze
PP29667. If ak > 0 (k = 1, 2, ..., n) , then
4P
cyclic
a23(a1+a2)(2a1+2a2+10a3)
≥ n7 − 9
49 ln�
12nQ a1+a2
a3
�.
Mihaly Bencze
PP29668. If xk > 0 (k = 1, 2, ..., n) , thenP 1
ch2xk(5+2ch2xk)≥ n
7 − 9n49 ln
�1 + 1
n
nPk=1
sh2xk
�.
Mihaly Bencze
PP29669. If xk ≥ 1 (k = 1, 2, ..., n) , then
nQk=1
�7xk
5+2xk
� 495
�nQ
k=1
xxk
k
�9
≥
7n�
k=1xk
5n+2n�
k=1
xk
495 �
1n
nPk=1
xk
�9n�
k=1xk
. If
0 < xk < 1 (k = 1, 2, ..., n) , then holds the reverse inequality.
Mihaly Bencze
Proposed Problems 1027
PP29670. In all triangle ABC holds
1). 15
Pln 7
2+5 sinA − 949
P ln(sinA)sinA ≥ 144
49
�s2+r2+4Rr
2sr − 3�
2). 15
Pln 7
2+5 sin2 A− 18
49
P ln(sinA)
sin2 A≥ 144
49
��s2+r2+Rr
2sr
�2− 4R
r − 3
�
Mihaly Bencze
PP29671. If x, y, z ≥ 1 then�Q 7x5+2x
� 15(Q
xx)949
�7�
x15+2
�
x
�3·��
x3
� 949
�
x≥Q
�7(x+y)5+x+y
� 25 �x+y
2
� 949
(x+y).
If 0 < x, y, z ≤ 1 then holds the reverse inequality.
Mihaly Bencze
PP29672. In all acute triangle ABC holds
1). 15
Pln 7
2+5 cosA − 949
P ln(cosA)cosA ≥ 144
49
�s2+r2−4R2
s2−(2R+r)2− 3
�
2). 15
Pln 7
2+5 cos2 A− 18
49
P ln(cosA)cos2 A
≥ 14449
��s2+r2−4R2
s2−(2R+r)2
�2− 8R(R+r)
s2−(2R+r)2− 3
�
Mihaly Bencze
PP29673. Prove thatnQ
k=1
k45k
(2k+5)49≥ e360n(n−1)
(7n·n!)49 .
Mihaly Bencze
PP29674. Prove thatnQ
k=1
k90k2
(2k2+5)49≥ e120n(n−1)(2n+5)
(7n(n!)2)49 .
Mihaly Bencze
PP29675. Prove thatnQ
k=1
(1+ 1k )
45(1+ 1k )
(7k+2)49≥ (n+1)720
(7n·(n+1)!)49.
Mihaly Bencze
PP29676. Prove thatnQ
k=1
k2+k+17k2+7k+1
�k2+k+1k2+k
� 45(k2+k+1)k2+k ≥ 7−ne
720nn+1 .
Mihaly Bencze
1028 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP29677. If x ≥ 1 then 495 ln 7x
2x+5 + 9x lnx ≥ 144 (x− 1). What happens ifx ∈ (0, 1)?
Mihaly Bencze
PP29678. In all triangle ABC holdsQ ( a+b
c )45(a+b)
c
(2a+2b+5c)49≥ e
360
�
s2+r2
Rr−8
�
(686s(s2+r2+2Rr))49.
Mihaly Bencze
PP29679. In all triangle ABC holdsQ
�
ra+rbrc
�
45(ra+rb)rc
(2ra+2rb+5rc)49 ≥ e
720
�
2(2R−r)r −3
�
(4s2R)49.
Mihaly Bencze
PP29680. If xk ≥ 1 (k = 1, 2, ..., n) , then
49P x1−1
x2−1 ln7x1
5+2x1+ 45
P�x1−1x2−1
�x1 lnx1 ≥ 720
�nP
k=1
xk − n
�.
Mihaly Bencze
PP29681. Prove thatn+1Pk=2
1
(49 ln 7k2k+5
+45k ln k)(49 ln 7k+72k+7
+45(k+1) ln(k+1))≤ n
158400(n+1) .
Mihaly Bencze
PP29682. In all triangle ABC holds1).
P2sinA (sinB + sinC)1−sinA ≤ 4
PsinA
2).P
2cosA (cosB + cosC)1−cosA ≤ 4P
cosA if ABC is acute
3).P
2sin2 A
�sin2B + sin2C
�cos2 A ≤ 4P
sin2A
4).P
2cos2 A
�cos2B + cos2C
�sin2 A ≤ 4P
cos2A
Mihaly Bencze
PP29683. In all quadrilateral ABCD holdsP(sinA)sinB (sinA+ sinB)sinC (sinA+ sinB + sinC)sinD ≥ 1.
Mihaly Bencze
PP29684. If x ∈ R then (sinx cosx)2 sin2 x + (sinx cosx)2 cos
2 x ≥ 1.
Mihaly Bencze
Proposed Problems 1029
PP29685. In all triangle ABC holds
1).P
(sinA+ sinB)sinC ≥ 2
2).P
(cosA+ cosB)cosC ≥ 2 if ABC is acute
Mihaly Bencze
PP29686. If ak ∈ (0, 1) (k = 1, 2, ..., n) thenPcyclic
aa21 (a1 + a2)a3 ... (a1 + a2 + ...+ an−1)
an ≥ 1.
Mihaly Bencze
PP29687. In all triangle ABC holds
1).P
(sinA)sinB (sinA+ sinB)sinC ≥ 1
2).P
(cosA)cosB (cosA+ cosB)cosC ≥ 1 if ABC is acute
Mihaly Bencze
PP29688. In all triangle ABC holds
1).P
(sinA+ sinB)sinC ≥ 2
2).P
(cosA+ cosB)cosC ≥ 2 if ABC is acute
Mihaly Bencze
PP29689. If ak ∈ (0, 1) (k = 1, 2, ..., n) , thenaa21 (a1 + a2)
a3 ... (a1 + a2 + ...+ an−1)an ≥ a1
a1+a2+...+an.
Mihaly Bencze
PP29690. If ak ∈�0, 1
n−1
�(k = 1, 2, ..., n) , then
Pcyclic
aa2+a3+...+an1 ≥ 1.
Mihaly Bencze
PP29691. If ak ∈ (0, 1) (k = 1, 2, ..., n) thenP(a1 + a2 + ...+ an−1)
an ≥ n− 1.
Mihaly Bencze
PP29692. If x > 1 thennP
k=1
xk−1−1xk+1
≤ n(n−1)(x−1)4x .
Mihaly Bencze
1030 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP29693. If a, b > 0 then bn+1−an+1
(n+1)(b−a) +√akbk
�bn−k+1−an−k+1
(n−k+1)(b−a)
�≤ an + bn, for
all n, k ∈ N.
Mihaly Bencze
PP29694. If 1 < a ≤ b thenbRa
(xn−1−1)dx(xn+1)(x−1) ≥
n−12 ln b
a .
Mihaly Bencze
PP29695. If xk > 1 (k = 1, 2, ..., n) and n ≥ 2 then
nPk=1
xk(xmk+1)
xm−1k
−1≥
2
�
n�
k=1xk
�2
(m−1)
�
n�
k=1
xk−n
� for all m ≥ 2.
Mihaly Bencze
PP29696. If λ, ai > 0 (i = 1, 2, ..., n) and k ≥ 2 thenP
cyclic
(λ+a1)k
λ2+λka2+k(k−1)
2a22
≥ nλk−2.
Mihaly Bencze
PP29697. In all triangle ABC holds
1).P ab
a2−ab+b2≥ s2+9r2+18Rr
2(s2−3r2−6Rr)
2).P rarb
r2a−rarb+r2b
≤ 6s2R(4R+r)2−2s2
− 1
Mihaly Bencze
PP29698. Determine all λk, µk > 0 (k = 1, 2, 3, 4) such that for all a, b, c > 0holds
�Paλ1bµ1
� �Paλ2bµ2
�≥
�Paλ3bµ3
� �Paλ4bµ4
�.
Mihaly Bencze
PP29699. In all acute triangle ABC holds
1).P (1+cosn A)(1−cosA) cos2 A
1−cosn−1 A≥ 2(s2+r2−4R2)
(n−1)(s2−(2R+r)2)
2).P (1+cos2n A) cos4 A sin2 A
1−cos2n−2 A≥ 2
n−1
��s2+r2−4R2
s2−(2R+r)2
�2− 8R(R+r)
s2−(2R+r)2
�for all
n ≥ 2
Mihaly Bencze
Proposed Problems 1031
PP29700. In all triangle ABC holds
1).P (1+sinn A)(1−sinA) sin2 A
1−sinn−1 A≥ s2+r2+4Rr
(n−1)sr
2).P (1+sin2n A) sin4 A cos2 A
1−sin2n−2 A≥ 2
n−1
��s2+r2+Rr
2sr
�2− 4R
r
�for all n ≥ 2
Mihaly Bencze
PP29701. In all acute triangle ABC holdsP tg2A−sin2 A
B2−sin2 B≥ 9.
Mihaly Bencze
PP29702. If xk ∈�0, π2
�(k = 1, 2, ..., n) then
nPk=1
tgxk ≥ 12
nPk=1
sin 2xk +nP
k=1
x3k.
Mihaly Bencze
PP29703. In all acute triangle ABC holdsP
A3 ≤ 2srs2−(2R+r)2
− srR2 .
Mihaly Bencze
PP29704. If x ∈�0, π2
�then tgx ≥ x3 + sin 2x
2 .
Mihaly Bencze
PP29705. In all acute triangle ABC holdsP
A�tg2A+ 2 sin2A
�≥ π3
3 .
Mihaly Bencze
PP29706. In all acute triangle ABC holdsP tg2A+2 sin2 A
A ≥ 3π.
Mihaly Bencze
PP29707. In all acute triangle ABC holdsP tg2A+2 sin2 A
B2 ≥ 9.
Mihaly Bencze
PP29708. In all acute triangle ABC holdss2−r(4R+r)
R2 +4s2r2−2(s2−r2−4Rr)(s2−r2−4Rr−4R2)
(s2−(2R+r)2)2 ≥ π2.
Mihaly Bencze
1032 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP29709. If xk ∈�π4 ,
π2
�(k = 1, 2, ..., n) , then
nPk=1
tg2xk + 2nP
k=1
sin2 xk ≥ 3nP
k=1
x2k + 2nP
k=1
xk.
Mihaly Bencze
PP29710. If a, b, c, d > 0 and x, y, z, t ∈ N ∗ then
��������
a −b 0 00 b −c 00 0 c −d1 1 1 1 + d
��������≥ abcd+ x+y+z+t
((ax)x(by)y(cz)z(dt)t)1
x+y+z+t.
Mihaly Bencze
PP29711. In all tetrahedron ABCD holds
��������
ha −hb 0 00 hb −hc 00 0 hc −hd1 1 1 1 + hd
��������+
��������
ra −rb 0 00 rb −rc 00 0 rc −rd1 1 1 1 + rd
��������=
=�1 + 1
r
�hahbhchd +
�1 + 2
r
�rarbrcrd.
Mihaly Bencze
PP29712. If x, y ∈ R then
��������
cosx+ cos y − cosx+ cos y − sinx+ sin y sinx+ sin y− cosx+ cos y cosx+ cos y sinx+ sin y − sinx+ sin y− sinx+ cos y sinx+ sin y cosx+ cos y − cosx+ cos ysinx+ sin y − sinx+ sin y − cosx+ cos y cosx+ cos y
��������=
= 16 cos 2x cos 2y.
Mihaly Bencze
PP29713. Solve in Z the equation x+y1+3z + y+z
1+3x + z+x1+3y = 3
2 .
Mihaly Bencze
Proposed Problems 1033
PP29714. If a, b, c > 0 and Δ1 =
������
1 a a3
1 b b3
1 c c3
������;
Δ2 =
������
a2 b2 c2
b2 + c2 c2 + a2 a2 + b2
bc ca ab
������then Δ1+Δ2
(b−a)(a−c)(b−c) ≥2√3(P
a)2 , for all
a 6= b 6= c.
Mihaly Bencze and Daniel Sitaru
PP29715. In all triangle ABC holds1).
P a2+b2
a4+6a2b2+b4≤ 1
8Rr
2).P r2a+r2
b
r4a+6r2ar2b+r4
b
≤ 4R+r4s2r
Mihaly Bencze
PP29716. In all triangle ABC holds1). s2+r2+4Rr
27sRr ≥P 2b−a(2a+b)(a+2b) +
16s
2). 427r ≥
P 2rb−ra(2ra+rb)(ra+2rb)
+ 13(4R+r)
Mihaly Bencze
PP29717. If a, b, c > 0 thenP 14a+1 + 3
P 12a+2b+1 ≤ 2
P 13a+b+1 + 2
P 1a+3b+1 + 12
P (a−b)4
(2a+2b+1)5.
Mihaly Bencze
PP29718. If a, b, c > 0 then13
P 1a+b+2c +
P 15(b+c)+2a ≥P 1
5a+4b+3c +P 1
5a+3b+4c .
Mihaly Bencze
PP29719. If a, b, c > 0 then�P a
c
� �48 +
�P ab
�3� ≥ 25�P a
b
�2.
Mihaly Bencze
PP29720. In all triangle ABC holds
1 ≥ (s2−r2−4Rr)(s2−(2R+r)2)2s2r2
+ 2R2−s2+(2R+r)2
s2−r(4R+r).
Mihaly Bencze
1034 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP29721. In all acute triangle ABC holds 3 ≥ 9+(�
ctgA)2
3+�
ctg2A.
Mihaly Bencze
PP29722. If a, b, c > 0 then 3310 ≤P 8a+7b
14a+13b +P 7a+8b
13a+14b ≤ 103 .
Mihaly Bencze
PP29723. If a, b, c > 0 thenP a2k−1+b2k−1
(1+c)2≥ (
�
a)2k−1
9k−1(1+(�
a)2).
Mihaly Bencze
PP29724. Determine all triangle ABC in which (P
cosActgB)2 ≥P cos2A.
Mihaly Bencze
PP29725. If xk ≥ 1 (k = 1, 2, ..., n) , then 3nP
k=1
x2k(3xk−1)
x2k+xk+1
≥
�
3n�
k=1
xk−n
�2
n+n�
k=1xk
.
Mihaly Bencze
PP29726. If xk ≥ 1 (k = 1, 2, ..., n) , then 3nP
k=1
x2k
x2k+xk+1
≥
�
3n�
k=1xk−n
�2
n�
k=1
x2k+2
n�
k=1
xk−n.
Mihaly Bencze
PP29727. Prove thatnP
k=1
k(3k−1)(k2+k+1)
≥ n3(n+1) .
Mihaly Bencze
PP29728. Prove that
nQk=1
(3k − 1)�k2 + k + 1
�≤ 3n (n!)2 (n+ 1)!.
Mihaly Bencze
PP29729. If x, y ∈�0, π2
�then x3y3 (sinx sin y + cos (x+ y)) ≤ sin3 x sin3 y.
Mihaly Bencze
Proposed Problems 1035
PP29730. In all acute triangle ABC holdsPtg2A+ 2
Psin2A ≥ 3
�A2 +B2 + C2
�.
Mihaly Bencze
PP29731. If xk ∈�0, π2
�(k = 1, 2, ..., n) , then sin
�nP
k=1
xk
�+
nQk=1
sinxk ≥ 0.
Mihaly Bencze
PP29732. If x, y, z ∈�0, π2
�, then
sinx sin y sin z + sin (x+ y + z) ≤P sinx sin3 y sin3 zy3z3
.
Mihaly Bencze
PP29733. If x, y, z ∈�0, π2
�, then
cosx cos y cos z − cos (x+ y + z) ≤P sinx sin y sin3 zz3
.
Mihaly Bencze
PP29734. If x ∈�0, π2
�, then 1
3 sin2 x cos2 x≥ x2 +
�π2 − x
�2.
Mihaly Bencze
PP29735. In all acute triangle ABC holdsP �tg2A+ 2 sin2A
� �tg2B + 2 sin2B
�≥ 9
PA2B2.
Mihaly Bencze
PP29736. In triangle ABC holds A ≤ B ≤ C ≤ π2 . Prove that
P Atg2A+2 sin2 A
≤ (π−B)2
4πAC .
Mihaly Bencze
PP29737. Prove thatnQ
k=1
(k − ln k) ≤ 1n!e
n(n−1)2 .
Mihaly Bencze
PP29738. If xi > 0 (i = 1, 2, ..., n) and k ∈ {1, 2, ..., n} , thenP
cyclic
�ex1+x2+...+xk−1 + (x2 + ...+ xk) ln (x1 + ...+ xk)
�≥ k2
n
�nP
i=1xi
�2
.
Mihaly Bencze
1036 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP29739. Prove thatnP
k=1
(k−ln k)(k+1−ln(k+1))e2k−1 ≤ n
n+1 .
Mihaly Bencze
PP29740. In all triangle ABC holdsP eA
A−lnA ≥ eπ.
Mihaly Bencze
PP29741. If ak > 0 (k = 1, 2, ..., n) , then
nQk=1
�eak + (1 + ak) ln (1 + ak)
�≥ 1 + n
snQ
k=1
ak
!2n
.
Mihaly Bencze
PP29742. If ak > 0 (k = 1, 2, ..., n) , thennQ
k=1
�eak(1+ak) +
�1 + ak + a2k
�ln�1 + ak + a2k
��≥
≥ 1 + n
snQ
k=1
ak + n
snQ
k=1
a2k
!2n
.
Mihaly Bencze
PP29743. Let A1A2...An be a convex polygon. Prove that
nPk=1
eAk + e ln
�nQ
k=1
Ak
�≥ (2 (n− 2)π − n) e.
Mihaly Bencze
PP29744. If ak > 0 (k = 1, 2, ..., n) , thennP
k=1
eak + ln
�nQ
k=1
(1 + ak)
�≥ n+ 2
nPk=1
ak.
Mihaly Bencze
PP29745. If xi > 0 (i = 1, 2, ..., n) and k ∈ N∗ thenP x1
x2+P
x1ek(x2−1) ≥ k+1
kk
k+1
nPi=1
xi.
Mihaly Bencze
Proposed Problems 1037
PP29746. If ak > 0 (k = 1, 2, ..., n) and An = 1n
nPk=1
ak and Gn = n
snQ
k=1
ak
then eGn−1 + lnGn + 2An ≤ eAn−1 + lnAn + 2Gn.
Mihaly Bencze
PP29747. If xi > 0 (i = 1, 2, ..., n) and k ∈ N∗ thenP
x1ekx2 +
P x11+x2
≥ k+1
kk
k+1
nPi=1
xi.
Mihaly Bencze
PP29748. Prove thatbRa
(ex−1+x lnx)(ex−1+lnx+1)dxx4+1
≥ 12 ln
1+b4
1+a4for all
1 ≤ a ≤ b.
Mihaly Bencze
PP29749. Prove thatnP
k=1
1
(ek+ln(k+1))(ek+1+ln(k+2))≤ n
3(2n+3) .
Mihaly Bencze
PP29750. Let A1A2...An be a convex polygon. Prove thatnP
k=1
eπAk
−1+ n lnπ ≥ n(n+2)
n−2 + ln
�nQ
k=1
Ak
�.
Mihaly Bencze
PP29751. If x ∈ R then esin2 x + ecos
2 x + ln��1 + sin2 x
� �1 + cos2 x
��≥ 4.
Mihaly Bencze
PP29752. If x ∈�0, π2
�then
esinx−1 + ecosx−1 + ln�(sinx)sinx (cosx)cosx
�≥ 1.
Mihaly Bencze
PP29753. In all triangle ABC holdsP
eA+BC + 3 lnπ ≥ 15 + ln (ABC) .
Mihaly Bencze
1038 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP29754. In all acute triangle ABC holds
1).P
e1
cosA−1 − ln
�s2−(2R+r)2
4R2
�+ 3 ≥ 2(s2+r2−4R2)
s2−(2R+r)2
2).P
etg2A − 2 ln
�s2−(2R+r)2
4R2
�+ 3 ≥ 2
�s2+r2−4R2
s2−(2R+r)2
�2− 16R(R+r)
s2−(2R+r)2
Mihaly Bencze
PP29755. If ak > 0 (k = 1, 2, ..., n) thennQ
k=1
(eak + ln (1 + ak) + 1) ≥ 2n
1 + n
snQ
k=1
ak
!n
.
Mihaly Bencze
PP29756. Prove thatbRa
(ex−1+x lnx)dxx6+1
≥ 13arctg
b3−a3
1+a3b3for all 0 < a ≤ b.
Mihaly Bencze
PP29757. In all triangle ABC holds1).
Pe
1sinA
−1 − ln�
sr2R2
�+ 3 ≥ s2+r2+4Rr
sr
2).P
ectg2A − 2 ln
�sr2R2
�+ 3 ≥ 2
�s2+r2+Rr
2sr
�2− 8R
r
Mihaly Bencze
PP29758. In all acute triangle ABC holdsPp(esinA−1 + sinA ln (sinA)) (ecosA−1 + cosA ln (cosA)) ≥ sr
R2 .
Mihaly Bencze
PP29759. In all triangle ABC holdsPr�
e− cos2 A2 + 2 sin2 A
2 ln�sin A
2
���e− sin2 A
2 + 2 cos2 A2 ln
�cos A
2
��≥
≥ s2−r(4R+r)8R2 .
Mihaly Bencze
PP29760. In all triangle ABC holds
1).P
esin3 A
2−1 + 3
Psin3 A
2 ln�sin A
2
�≥ (2R−r)((4R+r)2−3s2)+6Rr2
32R3
2).P
ecos3 A
2−1 + 3
Pcos3 A
2 ln�cos A
2
�≥ (4R+r)3−3s2(2R+r)
32R3
Mihaly Bencze
Proposed Problems 1039
PP29761. If x ∈ R thene− sin2 x + e− cos2 x + ln
�(sinx)2 sin
2 x (cosx)2 cos2 x�≥ 1− 2 sin2 x cos2 x.
Mihaly Bencze
PP29762. In all triangle ABC holds
1).P
e− cos2 A2 + 2
Psin2 A
2 ln�sin A
2
�≥ 8R2+r2−s2
8R2
2).P
e− sin2 A2 + 2
Pcos2 A
2 ln�cos A
2
�≥ (4R+r)2−s2
8R2
Mihaly Bencze
PP29763. Let ABCD be a convex quadrilateral andM1 ∈ (AB) , N1 ∈ (BC) ,M3 ∈ (CD) , N3 ∈ (DA) . The quadrilateralsM1M2M3M4 and N1N2N3N4 are the intersections of triangles AM3B andCM1D respective BN3C and AN1D. Determine all quadrilaterals ABCD forwhich Area [M1M2M3M4] +Area [N1N2N3N4] ≤ 1
2Area [ABCD] .
Mihaly Bencze
PP29764. Prove thatnP
k=1
1kp ≥ (
√4n+5−2)
2p
n2p−1 for all n ≥ 2 and p ∈ N.
Mihaly Bencze
PP29765. Prove thatnP
k=1
1√k≥ 2
m
�√m2n+m2 + 1−m
�for all n ≥ 2 and
m ≥ 2; n,m ∈ N.
Mihaly Bencze
PP29766. If x, y, z > 0 then (P√
x)4 ≥ 16
Pxy.
Mihaly Bencze
PP29767. Determine all ak > 0 (k = 1, 2, ..., n) for which
√n+ 1 ≤
s
an +
ran−1 +
q...+
pa2 +
√a1 ≤
√2n− 1 for all n ≥ 2.
Mihaly Bencze
1040 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP29768. If ak =
s
k +
rk − 1 +
q...+
p2 +
√1 then
n−13(2n+1) ≤
nPk=2
1a2ka2k+1
≤ n2(n+1) .
Mihaly Bencze
PP29769. If a, b, c > 0 then determine all xk, yk > 0 (k = 1, 2, 3) such thatP
cyclic
(ax1bx2cx3)1
x1+x2+x3 ≥ Pcyclic
(ay1by2cy3)1
y1+y2+y3 .
Mihaly Bencze
PP29770. If 0 < ak < m (k = 1, 2, ..., n) , then�nP
k=1
1m−ak
��nP
k=1
1ak
�≥
�2nm
�2.
Mihaly Bencze
PP29771. If 0 < λ < ak (k = 1, 2, ..., n) , thenP
cyclic
a21a2−λ ≥ 4nλ.
Mihaly Bencze
PP29772. If a, b, c, d > 0 and a 6= b 6= c 6= d then��������
1 1 1 1a b c da2 b2 c2 d21a2
1b2
1c2
1d2
��������≥ 4
4√
(abcd)5|(a− b) (a− c) (a− d) (b− c) (b− d) (c− d)| .
Mihaly Bencze
PP29773. In all triangle ABC holds
1).P a2b
a3+b≤ s
2).P r2ab
r3a+rb≤ 4R+r
2
Mihaly Bencze
PP29774. If ak > 0 (k = 1, 2, ..., n) thenP
cyclic
a21a2a31+a2
≤ 12
nPk=1
ak.
Mihaly Bencze
Proposed Problems 1041
PP29775. In all triangle ABC holds
1).P ab2
(a3+b)(b3+c)≤ 5s2+r2+4Rr
64sRr
2).P rar2b
(r3a+rb)(r3b+rc)≤ (4R+r)2+s2
16s2r
Mihaly Bencze
PP29776. If x, y, z ∈ C such that min {|x| , |y| , |z|} ≥ e then
(|x+ y|+ |y + z|+ |z + x|)|x|+|y|+|z|+|x+y+z| ≥≥ (|x|+ |y|+ |z|+ |x+ y + z|)|x+y|+|y+z|+|z+x|.If max {|x| , |y| , |z|} ≤ e then holds the reverse inequality.
Mihaly Bencze
PP29777. Prove thatnP
k=1
k−1k ≤ n1− 1
n .
Mihaly Bencze
PP29778. If x ≥ e then (1 + lnx)(1+lnx)x√xln x ≤
�x√xlnx
�(1+lnx)1+ln x
. If
x ∈ (0, e) then holds the reverse inequality.
Mihaly Bencze
PP29779. If n ≥ 3 then
�nP
k=2
logk (k + 1)
�n
< n
n�
k=2logk(k+1)
.
Mihaly Bencze
PP29780. If n ≥ 2 then�
4n
n+1
�(n+1)(2n)!≥
�(2n)!
(n!)2
�4n(n!)2
.
Mihaly Bencze
PP29781. If n ≥ 6 then�n2
�nn+1·n!3n < (n!)
n2n
6n <�n3
�nn+1·n!2n .
Mihaly Bencze
PP29782. If n ≥ 3 then�2n3
�3n2+3n+1 ≤�3n2 + 3n+ 1
�2n3
.
Mihaly Bencze
1042 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP29783. If e ≤ a1 ≤ a2 ≤ ... ≤ an then�a1+a2+...+ak
k
� (a1+a2+...+an)(ak+1+ak+2+...+an)n(n−k) >
>�a1+a2+...+an
n
� (a1+a2+...+ak)(ak+1+ak+2+...+an)k(n−k) >
>�ak+1+ak+2+...+an
n−k
� (a1+a2+...+ak)(a1+a2+...+an)
kn.
If 0 < a1 ≤ a2 ≤ ... ≤ an ≤ e then holds the reverse inequality.
Mihaly Bencze
PP29784. In all triangle ABC holdsP 1
2a+b +P b
a(a+2b) ≤s2+r2+4Rr
4sRr .
Mihaly Bencze
PP29785. In all triangle ABC holdsP
a�
rb2a+c +
rc2a+b
�≤ 4R+ r.
Mihaly Bencze
PP29786. In all triangle ABC holdsP 1
a
�1
2b+a + 12c+a
�≤ 1
2Rr .
Mihaly Bencze
PP29787. In all triangle ABC holds12 + 5
Psin4 A
2 + 5P
cos4 A2 ≥ 8
Psin6 A
2 + 8P
cos6 A2 .
Mihaly Bencze
PP29788. In all acute triangle ABC holds6 + 5
P 1sin2 A cos2 A
+ 2P 1
sinA + 2P 1
cosA ≤ 8P 1
sin3 A+ 8
P 1cos3 A
.
Mihaly Bencze
PP29789. In all triangle ABC holdsP
a
�1
(2a+b) sin2 C2
+ 1(2a+2) sin2 B
2
�≤ s2+r2−8Rr
r2.
Mihaly Bencze
PP29790. In all acute triangle ABC holds27 + 2
PsinA+ 2
PcosA ≥ 8
Psin3A+ 8
Pcos3A.
Mihaly Bencze
Proposed Problems 1043
PP29791. In all triangle ABC holds
6 + 2P 1
sin2 A2cos2 A
2
+ 5P 1
sin4 A2
+ 5P 1
cos4 A2
≤ 8P 1
sin6 A2
+ 8P 1
cos6 A2
.
Mihaly Bencze
PP29792. In all triangle ABC holds
10√3�s2−r(4R+r)
R2
�+ 12s
R + 9√3 ≥ 16s(s2−3r2−6Rr)
R3 .
Mihaly Bencze
PP29793. In all acute triangle ABC holds
27 + 4rR ≥ 10(s2−(2R+r)2)
R2 +16((R+r)((4R+r)2−3s2)+3R(s2−(2R+r)2))
R3 .
Mihaly Bencze
PP29794. In all triangle ABC holds
11 +10(8R2+r2−s2)
R2 ≥ 4rR +
16((2R−r)((4R+r)2−3s2)+6Rr2)R3 .
Mihaly Bencze
PP29795. In all triangle ABC holds
225 + 36rR +
30((4R+r)2−s2)R2 ≥ 16((4R+r)3−3s2(2R+r))
R3 .
Mihaly Bencze
PP29796. If xk ∈ [0, 1] (k = 1, 2, ..., n), then
94
nPk=1
11+2xk
≥ 2n
�nP
k=1
11+xk
�2
+
�
n�
k=1xk
�2
n�
k=1x2k+2
n�
k=1xk+n
.
Mihaly Bencze
PP29797. If xk ∈ [0, 1] (k = 1, 2, ..., n), then 94
nPk=1
1+2xk
2+x2k
≥
�
n+2n�
k=1xk
�2
n�
k=1x2k+2
n�
k=1xk+n
.
Mihaly Bencze
1044 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP29798. If xk ∈ (0, 1) (k = 1, 2, ..., n), then
n+ 274
nPk=1
x2k
(1−xk)2(1+2xk)
≥ 1n
�nP
k=1
1+2x2k
1−x2k
�2
.
Mihaly Bencze
PP29799. Determine all x, y ∈ R for which�5x2 + 2y + 1
� �5y2 + 2x+ 1
�≥ 64x3y3.
Mihaly Bencze
PP29800. If xk ∈ [0, 1] (k = 1, 2, ..., n) then 32
nPk=1
(2+x2k)
32
√1+2xk
≥
�
2n+n�
k=1
x2k
�2
n+n�
k=1xk
.
Mihaly Bencze
PP29801. If xk ∈ [0, 1] (k = 1, 2, ..., n) then
32
nPk=1
x2k
�
(1+2xk)(2+x2k)
≥
�
n�
k=1
xk
�2
n+n�
k=1xk
.
Mihaly Bencze
PP29802. If x ∈ R then
3√2 ≥
q�1 + 2 sin2 x
� �2 + sin4 x
�+p(1 + 2 cos2 x) (2 + cos4 x).
Mihaly Bencze
PP29803. Prove that 3n(n+2)2(n+1) ≥
nPk=1
√(k2+k+2)(2k4+4k3+2k2+1)
(k(k+1))32
.
Mihaly Bencze
PP29804. Prove thatnP
k=1
1√k(k+2)(2k2+1)
≥ 2n3(n+1) .
Mihaly Bencze
PP29805. In all triangle ABC holdsPq
(1 + 2 sinA)�2 + sin2A
�≤ 3(s+3R)
2R .
Mihaly Bencze
Proposed Problems 1045
PP29806. In all acute triangle ABC holds
Pp(1 + 2 cosA) (2 + cos2A) ≤ 3(4R+r)
2R .
Mihaly Bencze
PP29807. If xk ∈ [0, 1] (k = 1, 2, ..., n) then
nQk=1
(1 + 2xk)�2 + x2k
�≤
�32
�1 + 1
n
nPk=1
xk
��2n
.
Mihaly Bencze
PP29808. In all triangle ABC holds
1).5(s2−4Rr−r2)
4s2r3≥P 1
r3a
2).5(3s2−r2−4Rr)
16s2r3≥P 1
h3a
Mihaly Bencze
PP29809. If 0 ≤ a ≤ b ≤ 1 thenbRa
(x2+2)dx(x+1)2
≤ 98 ln
1+2b1+2a .
Mihaly Bencze
PP29810. Prove thatnP
k=1
2k2+1k(k+1)3
≤ 9n8(n+2) .
Mihaly Bencze
PP29811. In all triangle ABC holds
1). s2−3r2
4Rr ≥ 1 + 8�s2−r2−4Rrs2+r2+4Rr
�2
2). 1 + s2−12Rrr2
≥ 4�s2−2r2−8Rr
r(4R+r)
�2
3). 1− 12Rr + (4R+r)3
s2r≥ 4
��4R+r
s
�2 − 2�2
4). 4 +(2R−r)((4R+r)2−3s2)
2Rr2≥ 16
�8R2+r2−s2
s2+r2−8Rr
�
5). (4R+r)3−3s2r2s2R
≥ 2 + 16�(4R+r)2−s2
s2+(4R+r)2
�2
Mihaly Bencze
1046 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP29812. In all triangle ABC holds
1).P (a+b)2(a+c)2
b+c ≥ 16sRr(5s2+r2+4Rr)s2+r2+2Rr
2).P (ra+rb)
2(ra+rc)2
rb+rc≥ 2r
�6s2 + (4R+ r)2
�
Mihaly Bencze
PP29813. If a, b, c > 0 thenP(a+ b)2 (a+ c)2 ≥ 12abc (a+ b+ c) + 1
9
�P ��a2 + ab+ ac− bc���2 .
Mihaly Bencze
PP29814. If a, b, c > 0 then
12abc (a+ b+ c) +P �
a2 + ab+ ac− bc�2 ≤P
�a+ b+c
2
�4.
Mihaly Bencze
PP29815. In all triangle ABC holdsP (a+b)2(a+c)2
bc = 16s2 +P (b2+bc+ba−ca)
2
ca .
Mihaly Bencze
PP29816. In all scalene triangle ABC holdsP�
cos A2
sin B−C2
�2λ
≥ 2λ
3λ−1 for all
λ ≥ 1.
Mihaly Bencze
PP29817. In all scalene triangles ABC holdsP a2+c2
(a−b)2≥ 3.
Mihaly Bencze
PP29818. If a, b, c > 0, a 6= b 6= c thenP a2+c2+(a+b)2
(a−b)2≥ 5.
Mihaly Bencze
PP29819. If a, b, c ∈ R, a 6= b 6= c thennP
k=1
��a
a−b + k�2
+�
bb−c + k
�2+�
cc−a + k
�2�
≥ n(2n2+6n+7)3 .
Mihaly Bencze
Proposed Problems 1047
PP29820. If a, b, c > 0 then�√a2 + b2 + c2 + b
√3�(a+ c) ≤ 4
√3
3
�a2 + b2 + c2
�, and his permutations.
Mihaly Bencze
PP29821. In all tetrahedron ABCD holds1).
�1√ha
+ 1√hc
��1√hb
+ 1√hd
�≤ 1
r
2).�
1√ra
+ 1√rc
��1√rb
+ 1√rd
�≤ 2
r
Mihaly Bencze
PP29822. If a, b, c > 0 then�a2 + b2 + 1
� �b2 + c2 + 1
� �c2 + a2 + 1
�≥ abc
�2 + 3
√abc
�3.
Mihaly Bencze
PP29823. If ak > 0 (k = 1, 2, ..., n) , then
Qcyclic
�a21 + a22 + 1
�≥
nQk=1
ak
2 + n
snQ
k=1
ak
!n
.
Mihaly Bencze
PP29824. In all triangle ABC holds
1). 1√ra
+q
rrarc
+ 1√rb
≤ 2√r
2). 1√ha
+q
rhahc
+ 1√hb
≤ 2√r
Mihaly Bencze
PP29825. If a, b, c > 0 thenP√
a+ 2b+ 3c ≥2√2�√
a(c+a)+√
b(a+b)+√
c(b+c)�
√a+b+c
.
Mihaly Bencze
PP29826. If λ, ak > 0 (k = 1, 2, ..., n) thenP
cyclic
a31+λa2+λa3
≥ 3nλ+1
3
q�λ2
�2.
Mihaly Bencze
1048 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP29827. If xk > 0 (k = 1, 2, ..., n) , thenP x1
x2+P
x1ex2−1 ≥ 2
nPk=1
xk.
Mihaly Bencze
PP29828. If xk > 0 (k = 1, 2, ..., n), thenP x2
1x2
+P
x21ex2−1 ≥ 2
n
�nP
k=1
xk
�2
.
Mihaly Bencze
PP29829. If x ∈ R then etg2x + ectg
2x ≥ 3.
Mihaly Bencze
PP29830. Let f : [0, 1] → (0,+∞) be a continuous function, thennP
k=2
1k−1
1R0
kpf (x) dx ≤ n−1
n
1R0
f (x) dx+ lnn.
Mihaly Bencze
PP29831. If a, b, c, d > 0 and k, p ∈ N ∗ then�a2 + c2
� �b2 + d2
�≥�
ak+p√akbp + c
k+p√ckdp
�·�b
k+p√apbk + d
k+p√cpdk
�≥ (ac+ bd)2 .
Mihaly Bencze
PP29832. If xk > 0 (k = 1, 2, ..., n) and a, b > 0 then
min
�nP
k=1
x3k
ax2k+b
;nP
k=1
x3k
bx2k+a
�≥ 1
a+b
nPk=1
xk.
Mihaly Bencze
PP29833. If ai > 0 (i = 1, 2, ..., n) and p, k ∈ N thennP
i=1apn+ki ≥
�nP
i=1aki
��nQ
i=1ai
�p
.
Mihaly Bencze
PP29834. In all triangle ABC holds
1). 3√3
2 ≤Pq
rar+ra
≤ 36√s2r√
r+3√s2r
2). 3√3
2 ≤Pq
ha
r+ha≤ 3 6√R√
3√R+3√2s2r2
Mihaly Bencze
Proposed Problems 1049
PP29835. If a, b, c > 0 thenP 1
a2(b2(b2+c2)+c2(a+b)2)≤ 1
2a2b2c2.
Mihaly Bencze
PP29836. In all triangle ABC holds
1).P ra
√ra
ra√ra+2r
√rb
≤ 4R+r5r
2).P ha
√ha
ha
√ha+2r
√hb
≤ 3(s2+r2+4Rr)10R
Mihaly Bencze
PP29837. If a, b, c, d, e > 0 thenP a2
(b+c+d+e)(a−b)(a−c)(a−d)(a−e) ≤1
(a+b+c+d+e)2, for all a 6= b 6= c 6= d 6= e.
Mihaly Bencze
PP29838. If ak > 0 (k = 1, 2, ..., n) , thenQ �2 (1 + a1a2) + a21 − a22
�≤ 2n (1 +
Pa1a2)
n .
Mihaly Bencze
PP29839. In all triangle ABC holds
1). 3√3R√
s+3R≤P 1√
1+sinA≤ 3
6√2R2√
3√2R2+ 3√sr
2). 3√6R√
8R−r≤P 1
�
1+sin2 A2
≤ 36√16R2√
3√16R2+
3√r2
3). 3√6R√
10R+r≤P 1
�
1+cos2 A2
≤ 36√16R2√
3√16R2+
3√s2
Mihaly Bencze
PP29840. Prove thatPa8 +
Pa6b2 + 4
Pa4b4 + 2abc
Pa3b2 ≥ 4
Pa6c2 + 4a2b2c2
Pab.
Mihaly Bencze
PP29841. If 0 < ak ≤ 1 (k = 1, 2, ..., n) thenPcyclic
1√2(1+a1a2)+a21−a22
≥ n√n
�
�
�
�2
�
n+�
cyclic
a1a2
�
.
Mihaly Bencze
1050 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP29842. Determine all a, b, c > 0 for which 2abc+ 1 ≥ ab+ bc+ ca.
Mihaly Bencze
PP29843. If a, b, c > 0 then 3 + 5P
a2 ≥ 2P
a+ 4P
ab.
Mihaly Bencze
PP29844. If a, b, c > 0 thenP
a4 + 2P
a2b2 ≥ 2abcP
a+ 2P
ab3.
Mihaly Bencze
PP29845. If a, b, c > 0 thena2 + b2 + c2 ≥ ab+ bc+ ca+ (a− b) (b− c) + (b− c) (c− a) + (c− a) (a− b) .
Mihaly Bencze
PP29846. If a, b, c > 0 thenP
a2 ≥ 12
Pab+ 1
2
P ��a2 − bc�� .
Mihaly Bencze
PP29847. If k ≥ 1 and a, b, c > 0 then
3k−1P �
a2 − bc�2k ≥
��Pa2�2 − (
Pab)2
�k.
Mihaly Bencze
PP29848. In all triangle ABC holdsP
atg2A2 ≥ 2m where m is a positiveroot of the equation x3 + sx2 − 2sRr = 0.
Mihaly Bencze
PP29849. If a, b, c, d, e > 0 then√a2 + b2 − ab+
pb2 + c2 −
√2bc+
+pc2 + d2 −
√3ad+
qd2 + e2 − 1
2
�√6 +
√2�de ≥
pa2 + e2 +
√3ae.
Mihaly Bencze
PP29850. In triangle ABC holds 0 < A ≤ B ≤ C ≤ π2 . Prove that
sinCC ≤ s
πR ≤ sinAA .
Mihaly Bencze
Proposed Problems 1051
PP29851. If ak ∈�0, 5π12
�(k = 1, 2, ..., n) , then 1 ≤
n�
k=1tgak
n�
k=1
ak
≤ 12(2+√3)
5π .
Mihaly Bencze
PP29852. If 0 < m ≤ xk ≤ M (k = 1, 2, ..., n) and 0 < a < b then
a+m2
b+m2 ≤na+
n�
k=1a2k
nb+n�
k=1b2k
≤ a+M2
b+M2 .
Mihaly Bencze
PP29853. Complex numbers a, b, c are pairwise distinct and satisfy|a| = |b| = |c| = 1 and |a+ b+ c| ≤ 1. Prove that1). |a− b|+ |b− c|+ |c− a| ≥ |a+ b|+ |b+ c|+ |c+ a|2).
��a2 + bc��+
��b2 + ca��+
��c2 + ab�� ≥ |a+ b|+ |b+ c|+ |c+ a|
Mihaly Bencze
PP29854. Let ABC be a triangle in which 0 < m ≤ t1
sinu ≤ M where
t ∈ {a, b, c} , u ∈ {sinA, sinB, sinC} . Prove that m ≤ (4sRr)Rr ≤ M.
Mihaly Bencze
PP29855. If 0 < m ≤ ak ≤ M ≤ 1 (k = 1, 2, ..., n) , then
2m1+m2 ≤
n�
k=1ak
n+n�
k=1a2k
≤ 2M1+M2 .
Mihaly Bencze
PP29856. If ak ∈�0, π2
�(k = 1, 2, ..., n) then 2
π ≤n�
k=1sin ak
n�
k=1ak
≤ 1.
Mihaly Bencze
PP29857. If 0 < a1 ≤ a2 ≤ ... ≤ an ≤ π2 then sin an
an≤
n�
k=1sin ak
n�
k=1ak
≤ sin a1a1
.
Mihaly Bencze
1052 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP29858. If x > 0 then nn+1 ≤
nPk=1
�
k�
i=1ixi−1
��
k+1�
i=1ixi−1
�
�
k−1�
i=0(n−i)xi
��
k�
i=0(n−i)xi
� ≤ n(n+1)(n+2)3 .
Mihaly Bencze
PP29859. Prove that nn+1 ≤
nPk=1
�
k�
i=1ixi−1
��
k+1�
i=1ixi−1
�
�
k�
i=1i2xi−1
��
k+1�
i=1i2xi−1
� ≤ n for all x ≥ 0.
Mihaly Bencze
PP29860. Let ABC be a triangle in which 0 < A ≤ B ≤ C ≤ π2 . Prove that
tgA ≤ sR+r ≤ tgC.
Mihaly Bencze
PP29861. In all acute triangle ABC holds 2 ≤P sinA ≤ π.
Mihaly Bencze
PP29862. Let A1A2...An be a tangential polygon with inradius R andsemiperimeter s. Prove that1). s
R ≥ Pcyclic
ctgA1+A2+...+Ak
2k ≥ nctg (n−2)πn
2). sR ≥ k
(n−1k−1)
P1≤i1<...<ik≤n
ctgAi1
+Ai2+...+Aik
2k ≥ nctg (n−2)πn where
k ∈ {1, 2, ..., n}
Mihaly Bencze
PP29863. Let ABCD be a circumscribed quadrilateral around a circle ofradius R, and denote s his semiperimeter. Prove1). s
R ≥P ctg 2π−A6 ≥ 4
2). sR ≥ 2
3
�ctgA+B
4 + ctgA+C4 + ctgA+D
4 + ctgB+C4 + ctgB+D
4 + ctgC+D4
�≥ 4
Mihaly Bencze
PP29864. Let ABCD be a circumscribed quadrilateral around a circle ofradius R, and denote s his perimeter. Prove that 4s
R + 3P
ctg 2π−A6 ≥
≥P
ctgA2 +4
�ctgA+B
4 + ctgA+C4 + ctgA+D
4 + ctgB+C4 + ctgB+D
4 + ctgC+D4
�.
Mihaly Bencze
Proposed Problems 1053
PP29865. In all triangle ABC holds1
2R−r
�x5 +
Psin10 A
2
�+ 1
4R+r
�x5 +
Pcos10 A
2
�≥ 3x(x+1)
2R for all x > 0.
Mihaly Bencze
PP29866. Let ABCD be a cyclic quadrilateral. Prove thatPtg5A2 ≥P tgA
2 .
Mihaly Bencze
PP29867. In all tetrahedron ABCD holds1). r
P 1h5a≥ 1
hahbhchd
2). r2
P 1r5a
≥ 1rarbrcrd
Mihaly Bencze
PP29868. In all triangle ABC holds1). x5 +
P 1h5a≥ x(xr+1)R
2s2r3
2). x5 +P 1
r5a≥ x(xr+1)
s2r2for all x > 0
Mihaly Bencze
PP29869. If xk > 0 (k = 1, 2, ..., n) , thennP
k=1
x4k
3x3k+7
≥
�
n�
k=1xk
�4
3
�
n�
k=1xk
�3
+7n3
.
Mihaly Bencze
PP29870. If xk > 0 (k = 1, 2, ..., n) , then 10nP
k=1
x4k
3x3k+7
≥nP
k=1
xk.
Mihaly Bencze
PP29871. If a, b, c > 0 and λ ≥ 0 thenaλ
√a2 + 2bc+ bλ
√b2 + 2ca+ cλ
√c2 + 2ab ≤ (a+ b+ c)
√a2λ + b2λ + c2λ.
Mihaly Bencze
PP29872. If a, b, c > 0 and x, y ≥ 0 thenPcyclic
bxcy 3pa3 + bc (2a+ 3b+ 3c) ≤ (
Pa) 3p(P
a3x) (P
a3y).
Mihaly Bencze
1054 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP29873. Prove that1). sin π
4n+2 ≥ 12n 2). tg π
4n ≤ 1n
3). tg π4n − sin π
4n < 12n for all n ∈ N∗
Ionel Tudor
PP29874. Prove that sin 4π15 − sin π
15 =√15−
√3
4 . Solve in N∗ the equation
sin 4πn − sin π
n =√n−
√3
4 .
Ionel Tudor
PP29875. Solve in N∗ the equation tg4 πn − 4tg3 πn − 14tg2 πn − 4tg πn + 1 = 0
Ionel Tudor
PP29876. 1). Prove that 2m ≥ m for all m ∈ N.2). Solve in N the equation 2n − 3n = 2015.
Ionel Tudor
PP29877. If d > 1 dividet 2016, then the equation 22016 + 32016 = xd havenot solution in N.
Ionel Tudor
PP29878. Let be N = 52017. Determine the last 5 digits of N. Determinethe number of digits of N.
Ionel Tudor
PP29879. Let be N = 20177...7 which have the last 2017 digits equal with7. Determine bn in corp Z2017.
Ionel Tudor
PP29880. Determine all n,m ∈ N for which 4n+ 1 is prime and√m+
pm+
√m+ n is natural.
Ionel Tudor
PP29881. Solve in Z the equation�x3 − 1
�y2 +
�y3 − 1
�x2 = x
�y4 − y2 + 1
�+ y
�x4 − x2 + 1
�.
Mihaly Bencze
Proposed Problems 1055
PP29882. Solve in R the following system:
4�x51 + 1
�= 5 5
√4x2
4�x52 + 1
�= 5 5
√4x3
−−−−−−−−−4�x5n + 1
�= 5 5
√4x1
.
Mihaly Bencze and Ionel Tudor
PP29883. Let ABCD be a tetrahedron, and denote RA, RB, RC , RD thecircumradii of triangles BCD, CDA, ABD, ABC. Prove thatP
R2nA ≥ 1
33n·4n−1 (AB +AC +AD +BC +BD + CD)2n for all n ∈ N∗.
Mihaly Bencze and Daniel Sitaru
PP29884. Let ABCD be a tetrahedron and denote RA, RB, RC , RD
respective SA, SB, SC , SD the circumradii respective the semiperimeters oftriangles BCD, CDA, DAB, ABC. Prove that24316
P (RARBRC)2
SBSC≥ 3
√AB2 ·AC2 ·AD2 ·BC2 ·BD2 · CD2.
Mihaly Bencze
PP29885. Let ABCD be a tetrahedron and denote RA, RB, RC , RD
respective SA, SB, SC , SD the circumradii respective the semiperimeters of
triangles BCD, CDA, DAB, ABC. Prove thatP RA
SB≥ 1
6√3(P
SA)�P 1
SA
�.
Mihaly Bencze
PP29886. Determine a polynomial P ∈ Z [x] of minimal degree for whichx = 7
√8 + 7
√16 is a roote.
Ionel Tudor
PP29887. Solve in R the equation 4x5 − 5 5√4x+ 4 = 0.
Ionel Tudor
PP29888. Let ABCD be a tetrahedron, and denote RA, RB, RC , RD thecircumradii of triangles BCD, CDA, ABD, ABC. Prove that729 (RARBRCRD)
3 ≥ (AB ·AC ·AD ·BC ·BD · CD)2 .
Mihaly Bencze
1056 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP29889. Determine all n ∈ N∗ for which
1).nQ
k=1
(1 + Fk)1+Fk ≤ eFn+2−1
√eFnFn+1
2).nQ
k=1
(1 + Lk)1+Lk ≤ eLn+2−3
√eLnLn+1−2
Mihaly Bencze
PP29890. Determine all x, y > 1 for which(1 + lnx)1+ln y + (1 + ln y)1+lnx ≤ x
pyln y + y
√xlnx.
Mihaly Bencze
PP29891. If x, y > 1 then
arctg�(1+lnx)1+ln x
√xln x
�+ arctg
�(1+ln y)1+ln y√
yln y
�≤ arctg x+y
xy−1 .
Mihaly Bencze
PP29892. Solve in Z the equation 2x
3y+5z + 2y
3z+5x + 2z
3x+5y = 34 .
Mihaly Bencze
PP29893. If in triangle ABC we have A ≤ B ≤ C thenP (ln eπ
A )ln eπ
A
( πA)
12 ln π
A
≤�3(π−B)
2√AC
�2.
Mihaly Bencze
PP29894. Let A1A2...An be a convex polygon such that
α ≤ Ak ≤ β (k = 1, 2, ..., n) . Prove thatnP
k=1
�
ln eπAk
�ln eπAk
�
πAk
� 12 ln π
Ak
≤ n2(α+β)2
4(n−2)αβ .
Mihaly Bencze
PP29895. If xk > 0 (k = 1, 2, ..., n) , thennQ
k=1
2x1+...+2xk−1+xk+2xk+1+...+2xn
2x1+...+2xk−1+3xk+2xk+1+...+2xn≥ 1
3 .
Mihaly Bencze
Proposed Problems 1057
PP29896. If x, y, z > 0 then
Px2
�1 + x+ x2
�≥Pxy +
Pxy (x+ y − z) +
Pxy
�x2 + y2 − z2
�.
Mihaly Bencze
PP29897. In all triangle ABC holds
1). 5 (P√
a)�P√
ab�≥ (P√
a)3+ 18
√abc
2). 5�P√
ma
� �P√mamb
�≥
�P√ma
�3+ 18
√mambmc
3). 5
�Pqcos A
2
��Pqcos A
2 cos B2
�≥
�Pqcos A
2
�3
+ 9p
sR
Mihaly Bencze
PP29898. If a, b, c > 0 then(5P
a)�P
a2 + 3P
ab�≥ 4 (
Pa)3 + 9
Q(a+ b) .
Mihaly Bencze
PP29899. If a, b, c > 0 thenP��
ab
�4+�bc
�3+ c
a
�≥ 3 +
P a(a2c2+b4)b2c3
.
Mihaly Bencze
PP29900. Determinr all n, k ∈ N for whichP (ab)n
ck≥P ab
c for alla, b, c > 0.
Mihaly Bencze
PP29901. If x, y, z, a, b, c > 0 such that ax+ by = z, by + cz = x,
cz + ax = y then 16abc (P
ab)�P 1
a+b+2
�≤ 1.
Mihaly Bencze
PP29902. If xk > 0 (k = 1, 2, ..., n) and r ∈ N thennP
k=1
xn+rk ≥
�nQ
k=1
xk
��nP
k=1
xrk
�.
Mihaly Bencze
1058 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP29903. If ak ∈�0, π2
�(k = 1, 2, ..., n) , then
nPk=1
�kP
i=1sin ai
��kP
i=1cos ai
�≤ n(n+1)(2n+1)
12 sin
12n�
k=1
�
n(n+1)2
− k(k−1)2
�
ak
n(n+1)(2n+1)
.
Mihaly Bencze
PP29904. Let ABC be a triangle and denote the angle α,β, γ the angle of
triangle formed by ma,mb,mc. Prove thatP√
mamb cos γ ≤√22
Pma.
Mihaly Bencze
PP29905. In all acute triangle ABC holdsPapb (cosB cosC) ≤ 1
2
�s2 + r (4R+ r)
�.
Mihaly Bencze
PP29906. In all triangle ABC holdsP
cos A2 ≥P
qr2R + 2 sin2 A
2 .
Mihaly Bencze
PP29907. If ak ∈ R, ak + k > 0 (k = 1, 2, ..., n) andnP
k=1
kak ≤ n(n+1)(3n2−5n−4)24 , n ∈ N,n ≥ 3, then
nPk=1
kak+k ≥ 2.
Mihaly Bencze
PP29908. If ak > 0 (k = 1, 2, ..., n) and m, r ∈ N thenP
cyclic
am+r+11am2 ar3
≥nP
k=1
ak.
Mihaly Bencze
PP29909. If xk > 0 (k = 1, 2, ..., n) , thennP
k=1
xk
1+√xk
≤√n
n�
k=1xk
√n+
�
n�
k=1xk
.
Mihaly Bencze
PP29910. If a, b, c > 0 then 2√2P a
c ≥√2Pp
ab +
Pq�cb
�2+�ba
�2.
Mihaly Bencze
Proposed Problems 1059
PP29911. If ak > 0 (k = 1, 2, ..., n) , thenP a1(4a31+3a1a22+2a32)
a2(a1+2a2)(2a21+a22)≥ n.
Mihaly Bencze
PP29912. In all acute triangle ABC holdsP sinA cos3 B
(sin2 A−3 sin2 A sin2 B+2 sin2 B) sinB+
+2P cos2 A cosB
sinA(sinA cosB+sinC) ≥ 1.
Mihaly Bencze
PP29913. Prove thatnP
k=1
√k4+4k3+10k2+12k+5
(3k2+6k+5)(k2+2k+2)≤ n(3n+5)
6(n+1)(n+2) .
Mihaly Bencze
PP29914. If n ≥ 6 then��
n3
�n+ n! +
�n2
�n� ��n3
�n+
�n2
�n� ≥≥ 12
�n2
6
�n ��n3
�2n+ (n!)2 +
�n2
�2n�.
Mihaly Bencze
PP29915. Prove that
1).nP
k=1
Fk+2(Fk+Fk+2)2
Fk≥ 6FnFn+1 + 2F 2
n+1 + F 2n+2 − 3
2).nP
k=1
Lk+2(Lk+Lk+2)2
Lk≥ 6LnLn+1 + 2L2
n+1 + L2n+2 − 6
Mihaly Bencze
PP29916. Prove that
1).nP
k=1
(√Fk+
√Fk+2)
2(√Fk+
√Fk+1+
√Fk+2)
2
√FkFk+2
≥ 12 (2Fn+1 + 4Fn+2 − 9)
2).nP
k=1
(√Lk+
√Lk+2)
2(√Lk+
√Lk+1+
√Lk+2)
2
√LkLk+2
≥ 12 (2Ln+1 + 4Ln+2 − 9)
Mihaly Bencze
PP29917. Prove that 2nQ
k=1
�√k+
√k+1+
√k+2√
k+2−√k
�4≥ 81nn! ((n+ 1)!)2 (n+ 2)!.
Mihaly Bencze
1060 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP29918. Prove thatnP
k=1
qk+1
k3(3k2+6k+5)≥ n√
3(n+1).
Mihaly Bencze
PP29919. If 0 < a1 ≤ a2 ≤ ... ≤ an then√32
nPk=1
ak ≥P
cyclic
�
a1a3(a21+a22+a23)a1+a3
.
Mihaly Bencze
PP29920. If 0 < a1 ≤ a2 ≤ ... ≤ an then 1√3
nPk=1
ak ≥ Pcyclic
�
a1a3(a21+a22+a23)a1+a2+a3
.
Mihaly Bencze
PP29921. If a, b, c > 0 then
�q3abc�
ab +q
13
Pa��q
3abc�
ab +6√abc+
q13
Pa�≥
≥ 2√3 4
qabc
�
a�
ab
q3abc�
ab +3√abc+ 1
3
Pa.
Mihaly Bencze
PP29922. If 0 < a1 ≤ a2 ≤ ... ≤ an then
136
Pcyclic
(√a1+
√a3)
2(√a1+
√a2+
√a3)
2
√a1a3
≥nP
k=1
ak.
Mihaly Bencze
PP29923. If a, b, c > 0 then 2 (P
a)2 (P
ab)2 + 9Q
(a+ b) ≥≥ 8 (
Pa) (P
ab) + 3 (P
ab)3 + 3abc (P
a)3 .
Mihaly Bencze
PP29924. If a, b > 0 then
�q2aba+b +
4√ab+
qa+b2
��q2aba+b +
qa+b2
�≥
≥ 2√3 4√abq
2aba+b +
√ab+ a+b
2 .
Mihaly Bencze
Proposed Problems 1061
PP29925. If ak ≥ 1 (k = 1, 2, ..., n) , thennP
k=1
11+a2
k
≥ Pcyclic
11+a1a2
+
�
�
√(a1a2−1)|a1−a2|
�2
2�
(a21+1)(a22+1)(a1a2+1).
Mihaly Bencze
PP29926. If x, y, z, a, b, c, d > 0 thenp(a2 + b2)x2 − 2axy + 2bxz + y2 + z2+
+p(d2 + 1) y2 + (c2 + 1) z2 − 2 (c+ d) yz ≥
≥p
(a2 + b2)x2 + d2y2 + c2z2 + 2bdxy − 2acxz.
Mihaly Bencze
PP29927. In all triangle ABC holds1).
P ra√r2+r2a
≤ 3�
1+( rs )
43
2).P ha√
r2+h2a
≤ 3�
1+�
Rr
2s2
� 43
Mihaly Bencze
PP29928. If ak ≥√2 (k = 1, 2, ..., n) , then
nPk=1
1√1+a2
k
≥ n�
�
�
�1+ n
�
n�
k=1a2k
. If
0 < ak <√2 (k = 1, 2, ..., n) then holds the reverse inequality.
Mihaly Bencze
PP29929. In all triangle ABC holds1).
P 1√1+sin2 A
≤ 3�
1+�
sr
2R2
� 32
2).P 1√
1+cos2 A≤ 3
�
1+�
s2−(2R+r)2
4R2
�
23
3).P 1
�
1+sin4 A2
≤ 3�
1+( r4R)
43
4).P 1
�
1+cos4 A2
≤ 3�
1+( s4R)
43
Mihaly Bencze
1062 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP29930. If x, y, z ∈�0, π2
�then
Pcosx ≤P
qcosx cos ycos(x−y) .
Mihaly Bencze
PP29931. If a, b, c > 0 thenP (a+b)(a−b)2
(a2+b2)�
a+b+√
2(a2+b2)� ≤ 1 + 8(a+b)(b+c)(c+a)√
2(a2+b2)(b2+c2)(c2+a2).
Mihaly Bencze
PP29932. If ak ∈ [0, 1] (k = 1, 2, ..., n) , thennP
k=1
1√1+a2
k
≤ Pcyclic
1√1+a1a2
.
Mihaly Bencze
PP29933. If n,m, k ∈ N ∗ then 4e2P �
1 + 1n
�n �e−
�1 + 1
n
�n� ≤
≤ 1 + 16e3
�1 + 1
n
�2 �1 + 1
m
�2 �1 + 1
k
�2.
Mihaly Bencze
PP29934. If x ∈�0, π2
�then 1√
1+sin2 x+ 1√
1+cos2 x≤ 2√
1+sinx cosx.
Mihaly Bencze
PP29935. If a, b, c > 0 thenP √
ab(√a−
√b)
2
a+b ≤ 18 + 16abc
(a+b)(b+c)(c+a) .
Mihaly Bencze
PP29936. If x, y, z > 0 (x 6= y 6= z) then 8P (x−y)((x+y)(lnx−ln y)−2(x−y))
(x+y)2(lnx−ln y)2≤
≤ 1+128(x−y)(y−z)(z−x)(x+y)(y+z)(z+x)(lnx−ln y)(ln y−ln z)(ln z−lnx) .
Mihaly Bencze
PP29937. Denote Fk, Lk, Pk the kth Fibonacci, Lucas and Pell numbers.Prove that
nPk=1
Fk+1Fk+2
F 2k
+nP
k=1
Lk+1Lk+2
L2k
+ 2P Pk+1Pk+2
P 2k
≤ n4 + 4
nPk=1
FkLkPk
Fk+2Lk+2Pk+2.
Mihaly Bencze
Proposed Problems 1063
PP29938. In all triangle ABC holds1). 4r
P ra−rr2a
≤ 1 +�4rs
�2
2). 4rP ha−r
h2a
≤ 1 + 8Rr2
s2
Mihaly Bencze
PP29939. In all triangle ABC holds1).
Psin2A ≤ 1
4 +�rR
�2
2).P
cos2A ≤ 14 +
�sR
�2
Mihaly Bencze
PP29940. If a, b, c > 0 thenP (a+2b)(b+2c)
(a+3b+2c)2≤ 1
4 + 4(a+2b)(b+2c)(c+2a)(a+3b+2c)(b+3c+2a)(c+3a+2b) .
Mihaly Bencze
PP29941. In all triangle ABC holdsPsin2 2A ≤ 1 + 8
�sin2A sin2B sin2C + cos2A cos2B cos2C
�.
Mihaly Bencze
PP29942. In all triangle ABC holds
1).P (ab)2
(a+b)3≥ 1
16s3
�(s2+r2+4Rr)
2+8s2Rr
s2+r2+2Rr
�2
2).P (rarb)
2
(ra+rb)3 ≥ 1
2(4R+r)
�s2+r(4R+r)
4R
�2
Mihaly Bencze
PP29943. In all triangle ABC holds1).
P ab(a+b)2
≤ s2+r2+34Rr4(s2+r2+2Rr)
2).P rarb
(ra+rb)2 ≤ R+4r
4R
3).P sin2 A
2sin2 B
2
(sin2 A2+sin2 B
2 )2 ≤ 1
4 + 8Rr2
(2R−r)(s2+r2−8Rr)−2Rr2
4).P cos2 A
2cos2 B
2
(cos2 A2+cos2 B
2 )2 ≤ 1
4 + 8Rs2
(4R+r)3+s2(2R+r)
Mihaly Bencze
PP29944. In all triangle ABC holdsP (−a+b+c)(a−b+c)
c2≤ R+4r
R .
Mihaly Bencze
1064 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP29945. If a, b, c > 0 thenPq
aba+b ≤
r2 (P
a)�14 + 4abc
(a+b)(b+c)(c+a)
�.
Mihaly Bencze
PP29946. If a, b, c > 0 such that a3 + b3 + c3 ≥ nn√
(a+b)(b+c)(c+a)then
a+ b+ c ≥ 3√n+ 1 for all n ∈ N∗.
Mihaly Bencze
PP29947. If a, b, c ∈ [0,λ] where λ > 0 such that a+ b+ c = 2λ thena2 + b2 + c2 ≤ 2λ2.
Mihaly Bencze
PP29948. If a, b, c,λ > 0 such that (a+ b) (b+ c) (c+ a) ≥ λ6
a3+b3+c3then
a+ b+ c ≥ 2λ.
Mihaly Bencze
PP29949. If a, b, c > 0 such that (a+ b) (b+ c) (c+ a) ≥ nn√a3+b3+c3
then
a+ b+ c ≥ 3√n+ 1 for all n ∈ N∗.
Mihaly Bencze
PP29950. If ak > 0 (k = 1, 2, ..., n) such that λa1 ≥ a2 + a3 + ...+ an where
λ ∈ [1, 3) then
�nP
k=1
ak
��nP
k=1
1ak
�≥ (3− λ)n2 + 2 (λ− 3)n+ 4.
Mihaly Bencze
PP29951. If a, b, c > 0 and x, y, z, t > 0 such that x2P
a2 + 3y2 = tabc thenP 1a3+b3+z3
≤ t6xyz .
Mihaly Bencze
PP29952. In all triangle ABC holds
Q �a4 + 3
�≥ 4
�1 + 2
�s2 − r2 − 4Rr
��2.
Mihaly Bencze
Proposed Problems 1065
PP29953. In all triangle ABC holdsQ �
a2 + 3 sin2 A2
�+Q �
a2 + 3 cos2 A2
�≥
≥ r2
4R2
�1 +
P asin A
2
�2
+ s2
4R2
�1 +
P acos A
2
�2
.
Mihaly Bencze
PP29954. In all triangle ABC holdsQ �a2 + 6b2 + 6c2
�≥ 256 (
Pambmc +mambmc)
2 .
Mihaly Bencze
PP29955. If ak, bk > 0 (k = 1, 2, ..., n) then
nQk=1
�a2k + nb2k
�≥ n
nQ
k=1
bk +P
cyclic
a1b2b3...bn
!2
.
Mihaly Bencze
PP29956. If ak ∈ {z ∈ C| |z| = 1} (k = 1, 2, ..., n) , then�
2n�
k=1
ak
�2
n+�
cyclic
a1a2+P
cyclic
�a1−a21−a1a2
�2≥ n.
Mihaly Bencze
PP29957. In all triangle ABC holdsQ �
a2 + 3r2a�≥ 4 (
Parbrc + rarbrc)
2 .
Mihaly Bencze
PP29958. If 0 ≤ ak ≤�nk
�(k = 0, 1, ..., n) , then
nQk=1
ak +nQ
k=1
��nk
�− ak
�≤ (n!)n+1
�
n�
k=1k!
�2 .
Mihaly Bencze
PP29959. If x, y ∈ R then�
eix+eiy
1+ei(x+y)
�2+�
eix−eiy
1−ei(x+y)
�2≥ 1 + sin (x− y) .
Mihaly Bencze
PP29960. If xk ∈ R (k = 1, 2, ..., n) thennQ
k=1
sin2 xk +nQ
k=1
cos2 xk ≤ 1.
Mihaly Bencze
1066 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP29961. If 0 < ak ≤ bk ≤ ck (k = 1, 2, ..., n) thennQ
k=1
(bk − ak) (ck − ak) (ck − bk) ≥
≥
nY
k=1
bk −nY
k=1
ak
! nY
k=1
ck −nY
k=1
ak
! nY
k=1
ck −nY
k=1
bk
!.
Mihaly Bencze
PP29962. If x ∈�0, π2
�then
sin2 x(1+2 sinx)1+cosx + cos2 x(1+2 cosx)
1+sinx + 4 (1− sinx) cos2 x+ 4 (1− cosx) sin2 x ≤ 12.
Mihaly Bencze
PP29963. If x ∈ R then
sin4 x(1+2 sin2 x)1+cos2 x
+cos4 x(1+2 cos2 x)
1+sin2 x≤ 8 + 12 sin2 x cos2 x.
Mihaly Bencze
PP29964. If ln�√
2− 1�≤ xk ≤ ln
�√2 + 1
�(k = 1, 2, ..., n) , thenP
cyclic
��sh2x1 − sh2x2�� ≤ 2
Pcyclic
1(chx1chx2)
2 .
Mihaly Bencze
PP29965. If x ∈�0, π2
�then
(sinx+ cosx)6 (1− sinx cosx)4 ≥ (sinx cosx)5 .
Mihaly Bencze
PP29966. In all triangle ABC holdsP ��sin A−B
2
�� ≤P 1(1+cosA)(1+cosB) cos C
2
.
Mihaly Bencze
PP29967. In all triangle ABC holds1).
P |ra − rb| (r + ra) (r + rb) ≤ 2s2�s2 − 2r (4R+ r)
�
2).P
|ha − hb| (r + ha) (r + hb) ≤4s2r2(s2−r2−4Rr)
R2
Mihaly Bencze
Proposed Problems 1067
PP29968. If ak > e (k = 1, 2, ..., n) , then
1).nQ
k=1
(ln ak)a2k ≥
ln
n�
k=1a2k
n�
k=1ak
n�
k=1a2k
2).Q
cyclic
(ln a1)a1a2 ≥
ln
�
cyclic
a1a2
n�
k=1ak
�
cyclic
a1a2
Mihaly Bencze
PP29969. In all triangle ABC holdsP ��sin A−B
2
�� ≤P 1(1+sinA)(1+sinB) sin C
2
.
Mihaly Bencze
PP29970. In all triangle ABC holds
1).Q �
ln esinA
� 1sin2 A ≥
�ln
e� 1
sin2 A� 1
sinA
�� 1sin2 A
2).Q �
ln esinA
� 1sinB sinC ≥
�ln
e� 1
sinA sinB� 1
sinA
�� 1sinA sinB
Mihaly Bencze
PP29971. In all triangle ABC holds�ln e2
�
sinA3 sinA sinB sinC
��
sinA≤Q
�ln e2
sinB sinC
�sinA.
Mihaly Bencze
PP29972. If ak > e (k = 1, 2, ..., n) and λk > 0 (k = 1, 2, ..., n) , then
nQk=1
(ln ak)λkak ≥
ln
n�
k=1λkak
n�
k=1
λk
n�
k=1λkak
. If ak ∈ (1, e) (k = 1, 2, ..., n) , then
holds the reverse inequality.
Mihaly Bencze
PP29973. If a, b, c ∈�1e , 1
�, then
�ln a+b+c
3abc
�a+b+c ≥�ln 1
bc
�a �ln 1
ca
�b �ln 1
ab
�c.
Mihaly Bencze
1068 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP29974. If ak ∈�1e , 1
�(k = 1, 2, ..., n) , then
ln
n�
k=1ak
nn�
k=1
ak
n�
k=1ak
≥ Qcyclic
�ln 1
a2a3...an
�a1.
Mihaly Bencze
PP29975. In all acute triangle ABC holds
�ln e
�
cosA�
cos2 A
��
cosA≤Q
�ln e
cosA
�cosA.
Mihaly Bencze
PP29976. In all triangle ABC holds
1).�ln s2
r(4R+r)
� er ≤Q
�ln era
r
� 1rb
2).�ln s2
s2−2r(4R+r)
� er ≤Q
�ln era
r
� 1ra
Mihaly Bencze
PP29977. In all triangle ABC holds�ln e
�
sinA�
sinA sinB
��
sinA≤
≤�ln e
sinA
�sinB �ln e
sinB
�sinC �ln e
sinC
�sinA.
Mihaly Bencze
PP29978. In all acute triangle ABC holds�ln e
�
cosA�
cosA cosB
��
cosA≤
≤�ln e
cosA
�cosB �ln e
cosB
�cosC �ln e
cosC
�cosA.
Mihaly Bencze
PP29979. In all triangle ABC holds�ln e
�
sinA�
sin2 A
��
sinA≤
≤�ln e
sinA
�sinA �ln e
sinB
�sinB �ln e
sinC
�sinC.
Mihaly Bencze
Proposed Problems 1069
PP29980. If λk > 0, ak ∈�1e , 1
�(k = 1, 2, ..., n), then
ln
n�
k=1λk
n�
k=1λkak
n�
k=1λk
≥
≥nQ
k=1
�ln 1
ak
�λk
. If 0 < ak ≤ 1e (k = 1, 2, ..., n) , then holds the reverse
inequality.
Mihaly Bencze
PP29981. If ak ∈�1e , 1
�(k = 1, 2, ..., n) , then
ln
n�
k=1ak
�
cyclic
a1a2
n�
k=1
ak
≥
≥�ln 1
a1
�a2 �ln 1
a2
�a3...
�ln 1
an
�a1.
If ak ∈�0, 1e
�(k = 1, 2, ..., n) then holds the reverse inequality.
Mihaly Bencze
PP29982. Solve in Q the following equation: 2a
b+c +2b
c+a + 2c
a+b = 3
Mihaly Bencze
PP29983. If xk > 0 (k = 1, 2, ..., n) andnQ
k=1
xk = 1 then determine all λ ∈ R
for whichnP
k=1
1n−1+xλ
k
≤ 1.
Mihaly Bencze
PP29984. Prove thatnP
k=1
(−1)k+1 (nk)k ≥ 2n
n+1 .
Mihaly Bencze
PP29985. If xk > 0 (k = 1, 2, ..., n) andnP
k=1
xk (2xk + 1) = nn+1 then
nPk=1
(xk+1)2
25xk+1 ≥ n2
3(n+1) .
Mihaly Bencze
1070 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP29986. Prove thatnP
k=1
hk + 1
k+√k2+2
i= n(n+1)
2 when [·] denote the
integer part.
Mihaly Bencze
PP29987. In all triangle ABC holds
1).P a
b+c ≥ 5s2−6r2−24Rr3(s2−r2−4Rr)
2).P b+c−a
a ≥ 11s2−24r2−96Rr3(s2−2r2−8Rr)
3).P ra
rb+rc≥ 11(4R+r)2−24s2
6((4R+r)2−2s2)
4).P sin2 A
2
sin2 B2+sin2 C
2
≥ 44R2+5r2−6s2+4Rr3(8R2+r2−s2)
5).P cos2 A
2
cos2 B2+cos2 C
2
≥ 5(4R+r)2−6s2
3((4R+r)2−s2)
Mihaly Bencze
PP29988. Solve the equation x5 − 5x3 − 5x = a10+1a5
, when a ∈ R∗.
Mihaly Bencze
PP29989. In all triangle ABC holds
1).P sinA
sinB ≥ 1 + sr − sr
2R2
2).P�
sin A2
sin B2
�2
≥ 2− r2R −
�r4R
�2
3).P�
cos A2
cos B2
�2
≥ 3 + r2R −
�s4R
�2
Mihaly Bencze
PP29990. In all triangle ABC holdsPq
s−arb
≤p
sr .
Mihaly Bencze
PP29991. Solve the equation x4 − 4x2 =�a4−1a2
�2when a ∈ R\ {−1, 0, 1} .
Mihaly Bencze
Proposed Problems 1071
PP29992. Prove that
1).nP
k=1
k2(V kn )
2
n2k+2Lk≥ 1
Ln+2−3
2).nP
k=1
�kV k
n
nk+1Lk
�2≥ 1
LnLn+1−2 where Lk denote the kth Lucas number.
Mihaly Bencze
PP29993. In all triangle ABC holdsP 1
sin2 A(1+sinB sinC)≤ 3(s2+r2+4Rr−4sr)R2
2s2r2.
Mihaly Bencze
PP29994. If λ ∈ (−∞, 0] ∪ [1,+∞) thennP
k=1
�kV k
n
nk+1
�λ≥ 1
n2λ−1 .
Mihaly Bencze
PP29995. Prove that
1).nP
k=1
k2(V kn )
2
n2k+2Fk≥ 1
Fn+2−1
2).nP
k=1
�kV k
n
nk+1Fk
�2≥ 1
FnFn+1 where Fk denote the kth Fibonacci number.
Mihaly Bencze
PP29996. Let ABC be a triangle, determine all x, y ∈ R for which
cos (xA+ yB) cos (xB + yC) cos (xC + yA) ≥�2rR
�2.
Mihaly Bencze
PP29997. If ak > 1 (k = 1, 2, ...,m) then�
n�
k=1
(ak+1)
�m
−�
n�
k=1
ak
�m
�
n�
k=1ak
�m−2 n�
k=1(ak+1)
�
n�
k=1(ak+1)−
n�
k=1ak
� <
<mPj=1
�nQ
k=1
1+aik
1+(ak+1)i
�<
�
n�
k=1(ak+1)
�m
−�
n�
k=1ak
�m
�
n�
k=1ak
�m−1� n�
k=1(ak+1)−
n�
k=1ak
� .
Mihaly Bencze
1072 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP29998. Let ABCD be a convex quadrilateral in which A∡ = C∡ = 90◦
and M ∈ BD. Prove that 1AM2 + 1
CM2 ≤ 1AB2 + 1
BC2 + 1CD2 + 1
DA2 .
Mihaly Bencze
PP29999. If xk > 0 (k = 1, 2, ..., n) andnP
k=1
xk = 1 thennP
k=1
xk
3x2k+2
≥ n2
2n2+3.
Determine all a, b > 0 for whichnP
k=1
xk
ax2k+b
≥ n2
bn2+a.
Mihaly Bencze
PP30000. Let ABC be a triangle, D ∈ (BC) , the bisectors of angles B andC meet the cevian AD in points M and N. Prove that�MDMA
�2+�NDNA
�2 ≥ sin2 Asin2 B+sin2 C
.
Mihaly Bencze
PP30001. Prove thatnP
k=2
�
2
�
3√
...√k
k(k+1) < n−16(n+1) .
Mihaly Bencze
PP30002. If a, b ∈ R such thatab
�a2 − 3b2
� �3a2 − b2
�+ 4 = 2a
�a2 − 3b
�+ 2b
�3a2 − b2
�then exist x ∈ R
such that a =√2 sinx and b =
√2 cosx.
Mihaly Bencze
PP30003. If a, b > 0 then computeR (sinx−cosx)dx
2aex+b sinx .
Mihaly Bencze
PP30004. If x1 = 1 and xn+1 = λ+ 1xn
for all n ≥ 1 then determine all
λ ∈ R for whichnP
k=1
[xk] = 2n, when [·] denote the integer part.
Mihaly Bencze
Proposed Problems 1073
PP30005. In all tetrahedron ABCD holds1).
P 1(4r−ha)(4r−hb)(4r−hc)
= 0
2).P 1
(2r−ra)(2r−rb)(2r−rc)= 0
Mihaly Bencze
PP30006. If Ak ∈ M3 (R) (k = 1, 2, ..., n) and AiAj = AjAi
(i, j ∈ {1, 2, ..., n}) then
Pcyclic
det�A2
1 +A1A2 +A22
�+P
cyclic
det�A2
1 −A1A2 +A22
�≥ 3
nPk=1
(detAk)2 .
Mihaly Bencze
PP30007. Let ABC be a triangle and M ∈ Int (ABC) . Prove thatP 1
sin2 AMB2
≥ 2(�
MA)2
s2−r(4R+r).
Mihaly Bencze
PP30008. In all triangle ABC holds
cos2�aA+bB+cC
a+b+c
�+ cos2
�bA+cB+aC
a+b+c
�+ cos2
�cA+aB+cC
a+b+c
�≤ 3− s2
3R2 .
Mihaly Bencze
PP30009. In all triangle ABC holds
1). 2P√
ab ≤P�a�ba
� rhc + b
�ab
� rhc
�≤ 2
Pa
2). 2P√
ab ≤P�a�ba
� rrc + b
�ab
� rrc
�≤ 2
Pa
Mihaly Bencze
PP30010. If 0 < a < b and 0 < c < d < 1 then 2 (d− c)√ab ≤
≤ acbd−adbc
ln b−ln a
�1
ac+d−1 + 1bc+d−1
�≤ (a+ b) (d− c) .
Mihaly Bencze
PP30011. In all triangle ABC holds
1). 2P√
ab ≤P�a�ba
�sinC+ b
�ab
�sinC�≤ 2
Pa
2). 2P√
ab ≤P�a�ba
�sin2 C2 + b
�ab
�sin2 C2
�≤ 2
Pa
1074 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
3). 2P√
ab ≤P�a�ba
�cos2 C2 + b
�ab
�cos2 C2
�≤ 2
Pa
Mihaly Bencze
PP30012. In all acute triangle ABC holds 32sRr ≤≤Q
�a�ba
�cosC+ b
�ab
�cosC� ≤ 2s�s2 + r2 + 2Rr
�.
Mihaly Bencze
PP30013. If λ > 0 then in all triangle ABC holds1).
Pλ
rra ≤ λ+ 2 2).
Pλ
rha ≤ λ+ 2
Mihaly Bencze
PP30014. If λ > 0 then in all tetrahedron ABCD holds1).
Pλ
rha ≤ λ+ 3 2).
Pλ
r2ra ≤ λ+ 3
Mihaly Bencze
PP30015. If x, y, z > 0 and x+ y + z = 1 then8 + 27xyz ≥ 27 (xy + yz + zx) .
Mihaly Bencze
PP30016. In all triangle ABC holds64Q
(ma +mb)2 ≥ (
Pma)
3Q (3ma + 3mb −mc) .
Mihaly Bencze
PP30017. Prove that
�nP
k=1
�k+1k
� 1k+1
�= n where [·] denote the integer part.
Mihaly Bencze
PP30018. In all triangle ABC holds1).
P a2
b3+c3≥ 3s
2(s2−r2−4Rr)
2).P r2a
r3b+r3c
≥ 3(4R+r)
2((4R+r)2−2s2)
3).P sin4 A
2
sin6 B2+sin6 C
2
≥ 6R(2R−r)8R2+r2−s2
4).P cos4 A
2
cos6 B2+cos6 C
2
≥ 6R(4R+r)
(4R+r)2−s2
Mihaly Bencze
Proposed Problems 1075
PP30019. If a, b, c > 0 thenP 1
(a+b)(b+c) ≤9
4(ab+bc+ca) .
Mihaly Bencze
PP30020. If ak ∈ (0, 1) (k = 1, 2, ..., n) then
nPk=1
logak
�1n
nPi=1
ai
�≥
n3n�
k=1ak
�
n�
k=1ak
�n .
Mihaly Bencze
PP30021. In all triangle ABC holds
1).P
logsinAs3R ≥ 27Rr
2s2
2).P
logsin2 A2
2R−r6R ≥ 27r2R
2(2R−r)3
3).P
logcos2 A2
4R+r6R ≥ 27s2R
2(4R+r)3
Mihaly Bencze
PP30022. Compute the integer part of the expressionnP
k=1
3
q1 + 1
k! +�1k!
�2.
Mihaly Bencze
PP30023. In all triangle ABC holds
1).P 1
r2a(2r2a+rbrc)
≥ 13s2r2
2).P 1
h2a(2h
2a+hbhc)
≥ R6s2r3
Mihaly Bencze
PP30024. Let ABC be a triangle. Determine all cevians ca, cb, cc for whichP(a+ b) ca ≤ 2
√3�s2 − r2 − 4Rr
�.
Mihaly Bencze
PP30025. If a ∈ (0, 1) ∪ (1,+∞) then solve in R the following system
(a− 1)2 x21 +�a2 − 1
�x2 + 2a = 2ax3+1
(a− 1)2 x22 +�a2 − 1
�x3 + 2a = 2ax4+1
−−−−−−−−−−−−−−−−−(a− 1)2 x2n +
�a2 − 1
�x1 + 2a = 2ax2+1
.
Gyorgy Szollosy and Mihaly Bencze
1076 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP30026. If ak > 0 (k = 1, 2, ..., n) andnP
k=1
ak = n thenP a21
a1+a22≥ n
n−1 .
Mihaly Bencze
PP30027. Determine all A ∈ M3 (R) for which
A3 +A2 +A =
1 1 11 1 11 1 1
.
Mihaly Bencze
PP30028. In all triangle ABC holds1). 1√
2(s2−r2−4Rr)≤P a
bc+2(s2−r2−4Rr)≤ 1√
s2−r2−4Rr
2). 1√s2−2r2−8Rr
≤P s−a(s−b)(s−c)+s2−2r2−8Rr
≤q
2s2−2r2−8Rr
3). 1√(4R+r)2−2s2
≤P rarbrc+(4R+r)2−2s2
≤q
2(4R+r)2−2s2
4). 4√2R√
8R2+r2−s2≤P sin2 A
2
sin2 B2sin2 C
2+ 8R2+r2−s2
8R2
≤ 8R√8R2+r2−s2
5). 4√2R√
(4R+r)2−s2≤P cos2 A
2
cos2 B2cos2 C
2+
(4R+r)2−s2
8R2
≤ 8R√(4R+r)2−s2
Mihaly Bencze
PP30029. In all acute triangle ABC holdsP cosA
1+cosA cosB ≤ 52 − s2−(2R+r)2
4R2 .
Mihaly Bencze
PP30030. In all triangle ABC holdsP 1
r2a(r2+rbrc)
≤ 5s2−2r2
2s4r2.
Mihaly Bencze
PP30031. In all triangle ABC holdsP 1
h2a(r
2+hbhc)≤ (5s2−Rr)R
4s4r3.
Mihaly Bencze
PP30032. In all triangle ABC holds 1sr ≤P 1
ra(r+√rbrc)
≤√2
sr .
Mihaly Bencze
Proposed Problems 1077
PP30033. In all triangle ABC holds 1sr
qR2r ≤P 1
ha(r+hbhc)≤ 1
sr
qRr .
Mihaly Bencze
PP30034. In all triangle ABC holds
1).Pq
a+bc ≤ 3
2
qs2+r2+2Rr
2Rr
2).Pq
ab+c−a ≤ 3
qR2r
3).Pq
ra+rbrc
≤ 3q
Rr
4).Pr
sin2 A2+sin2 B
2
sin2 C2
≤ 32r
q(2R−r)(s2+r2−8Rr)−2Rr2
2R
5).Pr
cos2 A2+cos2 B
2
cos2 C2
≤ 32s
q(4R+r)3+s2(2R+r)
2R
Mihaly Bencze
PP30035. In all triangle ABC holds
1).P a3
b(a2+bc+b2)≥ 2(s2−r2−4Rr)
s2+r2+4Rr
2).P r3a
rb(r2a+rbrc+r2b)
≥ (4R+r)2−2s2
s2
3).P sin6 A
2
sin2 B2 (sin
4 A2+sin2 B
2sin2 C
2+sin4 B
2 )≥ 2(8R2+r2−s2)
s2+r2−8Rr
4).P cos6 A
2
cos2 B2 (cos4
A2+cos2 B
2cos2 C
2+cos4 B
2 )≥ 2((4R+r)2−s2)
s2+(4R+r)2
Mihaly Bencze
PP30036. If x > 0, n, k ∈ N, n ≥ 2, k ≥ 2 thennP
i=1
�k
qx+ i−1
n
�≤
�nk−1 [nx]
� 1k where [·] denote the integer part.
Mihaly Bencze
PP30037. If ak ≥ 1 (k = 1, 2, ..., n) , then
1).P
cyclic
a51+4
a42−2a32+2a22−a2+1≥ 5n
2).P
cyclic
(a51+4)(a52+4)(a43−2a33+2a23−a3+1)(a44−2a34+2a24−a4+1)
≥ 25n
Mihaly Bencze
1078 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP30038. In all triangle ABC holds
1).Pq
a+bc ≥
√2 +
qs2+r2+2Rr
2Rr
2).Pq
ab+c−a ≥ 1 +
q2Rr
3).Pq
ra+rbrc
≥√2 + 2
qRr
4).Pr
sin2 A2+sin2 B
2
sin2 C2
≥√2 + 1
r
q(2R−r)(s2+r2−8Rr)−2R2
2R
5).Pr
cos2 A2+cos2 B
2
cos2 C2
≥√2 + 1
s
q(4R+r)3+s2(2R+r)
2R
Mihaly Bencze
PP30039. Solve in R the following system:
x21 + n− 1 ≤ x2 + x3 + ...+ xnx22 + n− 1 ≤ x3 + x4 + ...+ x1−−−−−−−−−−−−−−x2n + n− 1 ≤ x1 + x2 + ...+ xn−1
.
Mihaly Bencze
PP30040. Let be zk ∈ C (k = 1, 2, ..., n) such that |z1| = |z2| = ... = |zn| .Prove that z1, z2, ..., zn are the affixum of a regular n-gon, if and only ifz1 |z2 − z3| = z2 |z3 − z4| = ... = zn |z1 − z2| .
Mihaly Bencze
PP30041. In all triangle ABC holdsP (4+sin5 A) sin4 B
sin5 A(sin4 B−2 sin3 B+2 sin2 B−sinB+1)≥ 15.
Mihaly Bencze
PP30042. If ak ≥ 1 (k = 1, 2, ..., n) thenP
cyclic
(a41−2a31+2a21−a1+1)(a2+1)
a51+4≤ 1
5
�n+
nPk=1
ak
�.
Mihaly Bencze
PP30043. Solve in R the following equation�arccos x1
2 + arccos 2x2� �
arccos x22 + arccos 2x3
�...
�arccos xn
2 + arccos 2x1�=
=�2π3
�2.
Mihaly Bencze
Proposed Problems 1079
PP30044. If ak ∈ [0, 1] (k = 1, 2, ..., n) , thennP
k=1
2+aka2k+ak+1
≥ n2 + 1
2
Pcyclic
2+a1a21+a1a2+a21a
22.
Mihaly Bencze
PP30045. ComputeR (3x4+6x3+x2+3x−1)dx
x8+2x6+3x4+6x2+12x+10.
Mihaly Bencze
PP30046. If a, b, c,λ > 0 thenP a4+(a2+b2)c2
b+λc ≥ (�
a)3
3(λ+1) .
Mihaly Bencze
PP30047. If ak > 0 (k = 1, 2, ..., n) thenP a1
an−11 +(n−1)a2a3...an
≤
�
n�
k=1ak
�2
n2n�
k=1ak
.
Mihaly Bencze
PP30048. If xk > 0 (k = 1, 2, ..., n) andnP
k=1
x2k = n then
nPk=1
x3k
xk+λ ≥nP
k=1
xk
xk+λ for all λ > 0.
Mihaly Bencze
PP30049. If x, y, z,λ > 0 thenP x3
√3x+λ
√x2+y2+z2
≥√
x2+y2+z2
3
P x√3x+λ
√x2+y2+z2
.
Mihaly Bencze
PP30050. If xk > 0 (k = 1, 2, ..., n) and λ > 0 andnP
k=1
x2k = n, thenP xk
xk+λ ≤ nλ+1 .
Mihaly Bencze
PP30051. The triangle ABC is equilateral if and only ifaλ+b(1−λ)
c = bλ+c(1−λ)a = cλ+a(−λ)
b when λ ∈ (0, 1).
Mihaly Bencze
1080 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP30052. If 0 < a < 1 then solve the following system:
xax2
1 = axa3
xax3
2 = axa4
−−−−−xa
x1
n = axa2
.
Mihaly Bencze
PP30053. In triangle ABC we have a ≤ b ≤ c. Prove that for all λ ≥ 1holds aλ + hλa ≤ bλ + hλb ≤ cλ + hλc .
Mihaly Bencze
PP30054. If a > b ≥ 2, a, b ∈ N then [a, b] + 2nP
k=1
[a+ k, b+ k] +
+ [a+ n− 1, b+ n+ 1] ≥ 2n√a−b
(a+ b+ n+ 1) .
Mihaly Bencze
PP30055. Denote a (n) the smallest natural number which don’t divide n.
Compute∞Pn=1
11+d2(n)
.
Mihaly Bencze
PP30056. In all triangle ABC holdsP 1
sinA cos A2
+ 4 ≥ 2P 1
cos A2cos π−A
4
.
Mihaly Bencze
PP30057. In all convex polygon A1A2...An holdsnP
k=1
1
sinAk cosAk2
≥ n
sin�
(n−2)πn
�
cos�
(n−2)π2n
� .
Mihaly Bencze
PP30058. If x, y, z ∈ (0, 1) thenP 1
x(1−x) ≥ 8.
Mihaly Bencze
PP30059. In all triangle ABC holdsP 1
sin2 A2cos2 A
2
≥ maxn
72R8R−r ;
72R10R+r
o.
Mihaly Bencze
Proposed Problems 1081
PP30060. If ak > 0 (k = 1, 2, ..., n) , thenP a1
a22(a2+a3)3 ≥ n3
(n−1)3(�
a1a2)2 .
Mihaly Bencze
PP30061. In all triangle ABC holds
1).P a
b+c ≥ 32 +
27(s2−7r2−10Rr)64s2
2).P b+c−a
a ≥ 3 +27(s2−13Rr+r2)
4s2
3).P ra
rb+rc≥ 3
2 +27((4R+r)3−s2(8R−r))
8(4R+r)3
Mihaly Bencze
PP30062. In all triangle ABC holds
1).P a
b ≥q
s2+r2
Rr − 5
2).P b+c−a
a−b+c ≥q
8Rr − 7
3).P ra
rb≥q
8Rr − 7
4).P�
sin A2
sin B2
�2
≥q
(2R−r)(s2+r2−8Rr)Rr2
− 9
5).P�
cos A2
cos B2
�2
≥q
(4R+r)(s2+(4R+r)2)Rs2
− 9
Mihaly Bencze
PP30063. In all triangle ABC in which A ≤ B ≤ C holds
1).�e sin2 A
2
�sin2 B2sin2 C
2 ≤�e sin2 B
2
�sin2 C2sin2 A
2 ≤�e sin2 C
2
�sin2 A2sin2 B
2
2).�e cos2 A
2
�cos2 B2cos2 C
2 ≥�e cos2 B
2
�cos2 C2cos2 A
2 ≥�e cos2 C
2
�cos2 A2cos2 B
2
Mihaly Bencze
PP30064. Let ABC be a triangle in which A ≤ B ≤ C ≤ π2 then
(e sinA)sinB sinC ≤ (e sinB)sinA sinC ≤ (e sinC)sinA sinB .
Mihaly Bencze
PP30065. Let ABC be a triangle in which A ≤ B ≤ C ≤ π2 then
(e cosA)cosB cosC ≥ (e cosB)cosA cosC ≥ (e cosC)cosA cosB .
Mihaly Bencze
1082 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP30066. If xk ≥ e (k = 1, 2, ..., n) then
�nP
k=1
xk
��nP
k=1
1xk
�n2
≤
≤ n2
�
n�
k=1xk
��
n�
k=1
1xk
�
. If xk ∈ (0, e) (k = 1, 2, ..., n) then holds the reverseinequality.
Mihaly Bencze
PP30067. If 0 < a1 ≤ a2 ≤ ... ≤ an ≤ e thenaa2a3...an1 ≥ aa1a3a4...an2 ≥ ...a
a1a2...an−1n .
Mihaly Bencze
PP30068. If xk, yk ≥ e (k = 1, 2, ..., n) and p∈ N∗, p ≥ 2 then
�nP
k=1
(xk + yk)p
��
n�
k=1
xpk
� 1p+
�
n�
k=1
ypk
� 1p
≤
≤ �
nPk=1
xpk
� 1p
+
�nP
k=1
ypk
� 1p
!p
�
n�
k=1
(xk+yk)p
� 1p
. If xk, yk ∈ (0, e)
(k = 1, 2, ..., n) then holds the reverse inequality.
Mihaly Bencze
PP30069. If ak, bk ≥ e (k = 1, 2, ..., n) then
�nP
k=1
akbk
�2
�
n�
k=1a2k
��
n�
k=1b2k
�
≥
≥��
nPk=1
a2k
��nP
k=1
b2k
���
n�
k=1akbk
�2
. If ak, bk ∈ (0, e) (k = 1, 2, ..., n) then
holds the reverse inequality.
Mihaly Bencze
PP30070. If a, b, c > 0 then9�a2 + b2 + c2
�2 ≥ 3P �
a2 + b2 − c2�2
+ 8 (P
ab)2 .
Mihaly Bencze
PP30071. If a, b, c ≥ e then (P
ab)�
a2 ≥�P
a2�� ab
. If a, b, c ∈ (0, 1) thenholds the reverse inequality.
Mihaly Bencze
Proposed Problems 1083
PP30072. If xk ∈ (0, 1) (k = 1, 2, ..., n) then
1). ne−2n ≤
nPk=1
x√xk
k ≤ n
2). ne−4n ≤Px
√x1
1 x√x2
2 ≤ n
3). e−2n
nPk=1
xk ≤Px1x√x2
2 ≤nP
k=1
xk
Mihaly Bencze
PP30073. If a, b, c > 0 then 16P
a2b2 ≤P�a2 + b2 − c2
�2+ 5
�Pa2�2
.
Mihaly Bencze
PP30074. If ak ≥ 1 (k = 1, 2, ..., n) thennP
i,j=1
(aiaj)2
ai+aj−1 ≥�
nPk=1
ak
�2
.
Mihaly Bencze
PP30075. If ak ≥ 1 (k = 1, 2, ..., n) thennP
i,j,k=1
(aiajak)2
aiaj+ajak+akai−ai−aj−ak+1 ≥�
nPk=1
ak
�3
.
Mihaly Bencze
PP30076. If x1, y1 ∈ N∗ such that x1 <√7y1 and
xn+1 + xnyn = yn+1
�x2n + 1
�for all n ≥ 1, then prove that
nPk=1
xk
yk< n
√7.
Mihaly Bencze
PP30077. If a, b ∈ N ∗ and a < b√7 then a4 + 2a2 + 2 < ab
�a2 + 1
�√7.
Mihaly Bencze
PP30078. If a, b, c ∈ N ∗ and ab+ bc+ ca ≥ 1358786 thena3+b3+c3
3 ≥ 2019 + abc.
Mihaly Bencze
PP30079. If k ∈ N ∗ thenP
i,j=1
(ij)k+1
(i+j−1)k≥ (n+1)2k+2
22k+2nk−2 .
Mihaly Bencze
1084 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP30080. In all triangle ABC holds
1).P
n
snQ
k=1
(a+ λk) ≤ 3 n
snQ
k=1
�2s3 + λk
�
2).P
n
snQ
k=1
(s− a+ λk) ≤ 3 n
snQ
k=1
�s3 + λk
�
3).P
n
snQ
k=1
(ra + λk) ≤ 3 n
snQ
k=1
�4R+r
3 + λk
�
4).P
n
snQ
k=1
�sin2 A
2 + λk
�≤ 3 n
snQ
k=1
�1− r
2R + λk
�
5).P
n
snQ
k=1
�cos2 A
2 + λk
�≤ 3 n
snQ
k=1
�2 + r
2R + λk
�for all λk > 0
(k = 1, 2, ..., n)
Mihaly Bencze
PP30081. If x, y, z,λk > 0 (k = 1, 2, ..., n) then
n
snQ
k=1
(x+ λk) + n
snQ
k=1
(y + λk)+
+ n
snQ
k=1
(z + λk) + 3 n
snQ
k=1
�x+y+z3 + λk
�≤
≤ 2 n
snQ
k=1
�x+y2 + λk
�+ 2 n
snQ
k=1
�y+z2 + λk
�+ 2 n
snQ
k=1
�z+x2 + λk
�.
Mihaly Bencze
PP30082. If λ > 0 and xk ∈�1λ ,λ
�(k = 1, 2, ..., n) then determine
max
����nQ
k=1
(1− xk)
���� .
Mihaly Bencze
PP30083. If xk ∈�12 , 2
�(k = 1, 2, ..., n), then
1).nP
k=1
(1− xk)2 ≤ 1
2
nPk=1
xk
2).
����nQ
k=1
(1− xk)
���� ≤s
12n
nQk=1
xk
Mihaly Bencze
Proposed Problems 1085
PP30084. In all tetrahedron ABCD holds
1).�λ+ 1
r
�4 ≥ 125�
1hahbhchd
+ λP 1
hahbhc
�
2).�λ+ 2
r
�4 ≥ 125�
1rarbrcrd
+ λP 1
rarbrc
�for all λ > 0
Mihaly Bencze
PP30085. In all triangle ABC holds
1).�x+ y + 1
r
�4 ≥ 125�x+ys2r
+ xy(4R+r)s2r
�
2).�x+ y + 1
r
�4 ≥ 125
�R(x+y)2s2r2
+xy(s2+r2+4Rr)
4s2r2
�for all x, y > 0
Mihaly Bencze
PP30086. In all triangle ABC holds1). 2
PctgA
2 ≤P 1sin A
2sin B
2
2). 2P
tgA2 ≤P 1
cos A2cos B
2
Mihaly Bencze
PP30087. If ak ≥ 1 (k = 1, 2, ..., n) thenP
cyclic
a1a2 ≥ 2nP
k=1
qa2k − 1.
Mihaly Bencze
PP30088. In all triangle ABC holds
������
1 1 1ra rb rcr4a − r3a r4b − r3b r4c − r3c
������=
= (rb − ra) (rc − ra) (rc − rb)�(4R+ r)2 − s2 − 4R− r
�.
Mihaly Bencze
PP30089. In all acute triangle ABC holds1). 2
PctgA ≤P 1
sinA sinB2). 2
PtgA ≤P 1
cosA cosB
Mihaly Bencze
1086 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP30090. In all triangle ABC holdsP�n√sin2A+
n√sin2B
�n+P�
n√cos2A+
n√cos2B
�n≤
≤ 3 +�P n
√sin2A
�n+�P n
√cos2A
�nfor all n ∈ N∗.
Mihaly Bencze
PP30091. In all triangle ABC holds
������
1 1 1a b ca4 − a3 b4 − b3 c4 − c3
������=
= (b− a) (c− a) (c− b)�3s2 − r2 − 12Rr − 2s
�.
Mihaly Bencze
PP30092. Let ABCD be a convex pentagon and denote M,N,K,L, T themidpoints of AC,AD,BD,BE,EC. Prove that3 (AB +BC + CD +DE + EA) >
> AC +BE + EC +BD +DA+ 2 (MK +NT + LN + LM) .
Mihaly Bencze
PP30093. In all triangle ABC holds3 +
p3 + 2
Pcos (A−B) ≥ 2
Pcos A−B
2 .
Mihaly Bencze
PP30094. In all triangle ABC holds
1).P �
sin A2 + sin B
2
�2+P �
cos A2 + cos B
2
�2 ≤ 3 +�P
sin A2
�2+
�Pcos A
2
�2
2).P �
sin2 A2 + sin2 B
2
�2+P �
cos2 A2 + cos2 B
2
�2 ≤≤ 12R2+4Rr+r2−s2
4R2 +�P
sin2 A2
�2+�P
cos2 A2
�2
Mihaly Bencze
PP30095. Prove that 2P
1≤i<j≤n
1ij > ln2 n− 4(n−2)
5(2n+1) for all n ≥ 2.
Mihaly Bencze
PP30096. Prove that 2P
1≤i<j≤n
1ij < (1 + lnn)2 − 830n−89
252(2n+1) for all n ≥ 1.
Mihaly Bencze
Proposed Problems 1087
PP30097. If ak > 0 (k = 1, 2, ..., n) thenP
cyclic
pa21 − a1a2 + a22 ≤
≤nP
k=1
ak +
snP
k=1
a2k −P
a1a2.
Mihaly Bencze
PP30098. In all acute triangle ABC holdsP(cosA+ cosB)2 +
P(sinA+ sinB)2 ≤
≤ 3 + (P
cosA)2 + (P
sinA)2 .
Mihaly Bencze
PP30099. In all triangle ABC holds1).
P√a2 − ab+ b2 ≤ 2s+
√s2 − 3r2 − 12Rr
2). 12
Pq3 (a− b)2 + c2 ≤ s+
√s2 − 3r2 − 12Rr
3).Pq
r2a − rarb + r2b ≤ 4R+ r +q(4R+ r)2 − 3s2
4).Pq
sin4 A2 − sin2 A
2 sin2 B2 + sin4 B
2 ≤ 2R−r2R +
√(4R+r)2−3s2
4R
5).Pq
cos4 A2 − cos2 A
2 cos2 B2 + cos4 B
2 ≤ 4R+r2R +
√(4R+r)2−3s2
4R
Mihaly Bencze
PP30100. Prove thatnP
k=1
�2�1 + 1
2 + ...+ 1k
�+ 1
k+1 + 1k+2 + ...+ 1
2k
�<
<
�2nPk=1
1k
�2
− 1 < 2536
nPk=1
�2�1 + 1
2 + ...+ 1k
�+ 1
k+1 + 1k+2 + ...+ 1
2k
�.
Mihaly Bencze
PP30101. Prove thatnP
k=1
�1 + 1
13
� �1 + 1
23
� �1 + 1
33
�...
�1 + 1
(k(k+1))3
�< n(3n+2)
n+1 .
Mihaly Bencze
PP30102. Prove thatnP
k=1
�1 + 1
22+ 1
32+ ...+ 1
(k(k+1))2
�< n(2n+1)
n+1 .
Mihaly Bencze
1088 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP30103. If Ak =kP
i=1
2i−12i then n
16(n+1) <nP
k=1
A2kA
2k+1 <
9n64(n+1) .
Mihaly Bencze
PP30104. Prove that
nPk=1
rk2+5k+3√
(k+2)(k+3)�
(k+2)√k+
√(k2−1)(k+3)
� ≤rnq
n(n+2)n+3 .
Mihaly Bencze
PP30105. If Ak =kQ
i=1
2i−12i then
nPk=1
A2kA
2k+1 <
n4(3n+4) .
Mihaly Bencze
PP30106. If x, y, z > 0 thenP x
1+√x+
√3�
x√3+√
�
x≤P
√2(x+y)√2+
√x+y
.
Mihaly Bencze
PP30107. Prove that 2P
1≤i<j≤n
1√ij
≥ 4n2
(1+√n)
2 − ln en.
Mihaly Bencze
PP30108. In all triangle ABC holds
1).P a
1+√a≤ 2
√3s√
3+√2s
2).P s−a
1+√s−a
≤√3s√
3+√s
3).P ra
1+√ra
≤√3(4R+r)√3+
√4R+r
4).P sin2 A
2
1+sin A2
≤√6(2R−r)√
6R+√
R(2R−r)
5).P sin4 A
2
1+sin2 A2
≤√3(8R2+r2−s2)
2R�
4R√3+√
2(8R2+r2−s2)�
6).P cos2 A
2
1+cos A2
≤√3(4R+r)
2√3R+
√2R(4R+r)
7).P cos4 A
2
1+cos2 A2
≤√3((4R+r)2−s2)
2R�
4√3+
�
2R((4R+r)2−s2)�
Mihaly Bencze
Proposed Problems 1089
PP30109. In all triangle ABC holdsP
wa ≤r
2(4R+r)R
P�aba+b
�2≤ 1
2
q�4 + r
R
�(3s2 − r2 − 4Rr).
Mihaly Bencze
PP30110. In all triangle ABC holdsPq
b+cbc wa ≤
r(4R+r)((s2+r2+4Rr)2+8s2Rr)
sR(s2+r2+2Rr).
Mihaly Bencze
PP30111. In all triangle ABC holdsP
wa ≤ 2sqP ab
(a+b)2≤ s
√3.
Mihaly Bencze
PP30112. In all triangle ABC holds
P√b+ cwa ≤
r2s((s2+r2+4Rr)2+8s2Rr)
s2+r2+2Rr.
Mihaly Bencze
PP30113. In all triangle ABC holdsP 2(s−a)(s−b)+c(s−c)√
ab≤ 2s.
Mihaly Bencze
PP30114. In all triangle ABC holds Rr ≥ s2+r2−2Rr
6Rr ≥ s2+r2+10Rr12Rr ≥ 2. A
refinement of Euler’s inequality.
Mihaly Bencze
PP30115. In all tetrahedron ABCD holds1). 4 + r2
4
�1
rarb+ 1
rarc+ 1
rard+ 1
rbrc+ 1
rbrd+ 1
rcrd
�+ r4
8rarbrcrd≥
≥ 2 + r3
8 (ra + rb + rc + rd)
4). 4 + r2�
1hahb
+ 1hahc
+ 1hahd
+ 1hbhc
+ 1hbhd
+ 1hchd
�≥
≥ 2 + r3
hahbhchd(ha + hb + hc + hd)
Mihaly Bencze
PP30116. In all triangle ABC holdsP√
a+ b− c ≤ (3+2λ2)√2s
2λ for allλ > 0.
Mihaly Bencze
1090 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP30117. Let A1A2...An be a convex polygon. Prove thatP
cyclic
q(a1 + a2 + ...+ an−1)
2 − a2n ≤pn (n− 2)
nPk=1
ak.
Mihaly Bencze
PP30118. In all acute triangle ABC holds1).
PHA2 ≥ 4
�s2 + r2 − 4Rr
�
2).P HA
HB+HC ≥ (R+r)2
2(s2+r2−4R2)
Mihaly Bencze
PP30119. In all triangle ABC holds1).
Paw2
c ≤ 2s3
3
2).P
cwa ≤√2s√s2 − r2 − 4Rr
Mihaly Bencze
PP30120. In all triangle ABC holds 2P a(b+c)
b+c−a + 8s ≥P (a+b)(a+b+2c)c .
Mihaly Bencze
PP30121. In all triangle ABC holds
1).P�
a(b+c)
(b+c−a) sin A2
�2
≥ 32s2R2R−r
2).P�
a(b+c)
(b+c−a) cos A2
�2
≥ 32s2R4R+r
Mihaly Bencze
PP30122. In all triangle ABC holdsP�
a(b+c)(b+c−a)ma
�2≥ 32s2
3(s2−r2−4Rr).
Mihaly Bencze
PP30123. In all triangle ABC holdsP�
a(b+c)(b+c−a)ra
�2≥ 16s2
(4R+r)2−2s2.
Mihaly Bencze
PP30124. In triangle ABC the centroid belongs to the incircle. Prove that
12s2R+P
r3a =�
s2
4R
�3.
Mihaly Bencze
Proposed Problems 1091
PP30125. In triangle ABC the centroid belongs to the incircle. Prove that1
a2b2+ 1
b2c2+ 1
c2a2= 3
32R2r2.
Mihaly Bencze
PP30126. Let Ta, Tb, Tc be the circumradii of triangles BGC, CGA, AGB.
Prove that 54s2r2TaTbTc ≤�s2 − r2 − 4Rr
�3.
Mihaly Bencze
PP30127. Denote N the Nagel’s point in triangle ABC. Prove that�Psin2 A
2
� �Pcos2 A
2
�= (2NO+3r)(4NO+9r)
4(NO+2r)2.
Mihaly Bencze
PP30128. In all triangle ABC holdss2−r(4R+r)
Rr ≤P cos A2
sin B2sin C
2
≤ 31−1λ
�Paλ
� 1λ for all λ ∈ (−∞, 0] ∪ [1,+∞) .
Mihaly Bencze
PP30129. In all triangle ABC holds1).
Pm2
ar2a ≥ (4R+ r) s2r
2).P 1
m2a≤ 4R+r
s2r
3).P m2
a
ra≥ s2−2r(4R+r)
r
4).P m2
a
rar2c≥ (4R+r)2−2s2
s2r
5).P
m4a ≥ s2
�s2 − 2r (4R+ r)
�
6).Q �
m2a +m2
b
�≥ 4s4Rr
7).P m2
a+m2b
(ra+rb)2 ≥ 1
2
�4R+r
s
�2
Mihaly Bencze
PP30130. In all triangle ABC holds
1).P m2
a
rb≥ 4R+ r
2).P m2
a
hb≥ 3 3
qR2r
3).P m6
a
r3b
≥ (4R+ r)3 − 12s2R
Mihaly Bencze
1092 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP30131. In all triangle ABC holds
Pmamb ≤ 3
r3�s2r2 + 1
32
P(a2 − b2)2
�.
Mihaly Bencze
PP30132. In all triangle ABC holds
1).P ra
rb+rc= (4R+r)3+(8R+r)s2
4s2R
2). (4R+r)2
2s2≤P ra
rb+rc≤ R
r − 12
Mihaly Bencze
PP30133. In all triangle ABC holds 33√s2r ≤Pma ≤ 3
2
p2 (s2 − r2 − 4Rr).
Mihaly Bencze
PP30134. In all triangle ABC holdsQ
sin A+B4 ≥ r
4R .
Mihaly Bencze
PP30135. In all triangle ABC holds 3 3
q�r4R
�2 ≤P 1AI·BI ≤ 1
r2
Psin A+B
4 .
Mihaly Bencze
PP30136. Let A1B1C1 denote the Morley triangle of triangle ABC. DenoteRm and r the circumradii of triangle A1B1C1 and the inradi of triangle
ABC. Prove that Rm ≤ 2√3r3 .
Mihaly Bencze
PP30137. In all triangle ABC holds1).
P a2
bmb≥ 2
√3
2).P b2+λc2
ama≥ 2 (λ+ 1)
√3 for all λ ≥ 0
3).P a2+xb2+yc2
ama≥ 2 (x+ y + 1)
√3 for all x, y ≥ 0
Mihaly Bencze
PP30138. In all triangle ABC holdsP
3p
a2b2m2a ≤ 8
3 3√3
�s2 − r2 − 4Rr
�.
Mihaly Bencze
Proposed Problems 1093
PP30139. In all triangle ABC holdsQ�√3ama +
2s2
3
�≤ 8
�s2 − r2 − 4Rr
�3.
Mihaly Bencze
PP30140. In all triangle ABC holdsP (b+c) cos A
2wa
≥ 8s2
s2+r2+4Rr.
Mihaly Bencze
PP30141. Denote AD, BE, CF the Gergonne’s cevians in triangle ABC.
Prove that AD2 +BE2 + CF 2 =2(s2+r2)r
R + 2s2 + 7r2 − 4Rr.
Mihaly Bencze
PP30142. In all triangle ABC holds 2R(4R+r)2
2R−r ≤P�
rasin A
2
�2
≤ s2 +�9R2
�2.
Mihaly Bencze
PP30143. Denote AD, BE, CF the Gergonne’s cevians in triangle ABC.Prove that
AD ·ra+BE ·rb+CF ·rc ≤r�
(4R+ r)2 − s2��
2(s2+r2)rR + 2s2 + 7r2 − 4Rr
�.
Mihaly Bencze
PP30144. In all triangle ABC holds
P(ahb + bha) ≤ s2−3r2
2 − 6Rr +(s2+r2+4Rr)
2−24s2Rr
2R2 .
Mihaly Bencze
PP30145. In all triangle ABC holdsP ab+4sr
b2(4sRr+4(a+c)sr+ac2)≥ 3
8sRr .
Mihaly Bencze
PP30146. In all triangle ABC holdsP bma+2sr
b2((mb+mc)ac+2(a+c)sr)≥ 3
8sRr .
Mihaly Bencze
1094 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP30147. In all triangle ABC holds
1).P AI2
b+c =(s2+r2+4Rr)
2−4Rr(3s2+r(4R+r))2s(s2+r2+2Rr)
2).P
AI ≤p3 (s2 + r2 − 8Rr)
Mihaly Bencze
PP30148. Prove that
2nP
k=1
(ln(k+1))p+1−(ln k)p+1
(2k+1) ln(1+ 1k )
< (ln (n+ 1))p+1 < 12
nPk=1
(2k+1)((ln(k+1))p+1−(ln k)p+1)k(k+1)
for all p ∈ N.
Mihaly Bencze
PP30149. Solve in Z the equation 1x+yz + 1
y+zx + 1z+xy = 19
35 .
Mihaly Bencze
PP30150. If x ≥ 1 then�1 + 1
x
�x ≥ e4(x2+1)arctgx
πx(2x+1) .
Mihaly Bencze
PP30151. Prove that 2nP
k=1
ln2 k+ln k ln(k+1)+ln2(k+1)2k+1 < ln3 (n+ 1) <
< 12
nPk=1
(2k+1)(ln2 k+ln k ln(k+1)+ln2(k+1))k(k+1) .
Mihaly Bencze
PP30152. If 0 < x ≤ y then ln y(x+1)x(y+1) <
4(y−x)(2x+1)(2y+1) .
Mihaly Bencze
PP30153. If 1 ≤ a ≤ b then 2b+12a+1 ≤ aa(b+1)b+1
bb(a+1)a+1 ≤q
b(b+1)a(a+1) .
Mihaly Bencze
PP30154. Prove that π4−902985840 <
∞Pk=1
�e1+
12+...+ 1
k−γ − k − 1
2
�4< π4
2985840
where γ = 0, 57... denote the Euler constant.
Mihaly Bencze
Proposed Problems 1095
PP30155. If 0 < x ≤ y then (y−x)(2xy+x+y+1)2xy(x+1)(y+1) ≥ ln y(x+1)
z(y+1) .
Mihaly Bencze
PP30156. If Hk = 1 + 12 + ...+ 1
k then
� ∞Pn=1
Hn
n·2n
�2
= 512
� ∞Pn=0
H2n
(n+1)2
�.
Mihaly Bencze
PP30157. If ξ denote the Reiemann zeta function, then� ∞Pn=1
(ξ (2n)− 1)
�2
+ 7
� ∞Pn=1
(ξ (2n+ 1)− 1)
�2
= 1.
Mihaly Bencze
PP30158. Prove that�36n2 + 12n+ 1
�� nPk=0
�6n+36k
�B6k
�·�
nPk=0
�6n+56k+2
�B6k+2
�=
=�36n2 + 48n+ 15
�� nPk=0
�6n+16k−2
�B6k−2
�2
where Bk denote the kth Bernoulli
numbers.
Mihaly Bencze
PP30159. If γ = 0, 57... denote the Euler constant thenn(n+2)
2 + 124 ln
n+1e <
nPk=1
e1+12+...+ 1
k−γ < n(n+2)
2 + 124 lnne for all n ≥ 1.
Mihaly Bencze
PP30160. Prove that π2−63456 <
∞Pk=1
�e1+
12+...+ 1
k−γ − k − 1
2
�2< π2
3456 where
γ = 0, 57... denote the Euler constant.
Mihaly Bencze
PP30161. Prove that P
1≤i1≤n+1
1i1
!λ
+
P
1≤i1<i2≤n+1
1i1i2
!λ
+ ...+
P
1≤i1<i2<...<in≤n+1
1i1i2...in
!λ
≤
≤ n�n+ 1− 1
(n+1)!
�λfor all λ ∈ [0, 1] .
Mihaly Bencze
1096 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP30162. Prove thatnP
k=0
(nk)(2n−1
k )≥ 2
(n+1)√n+1
.
Mihaly Bencze
PP30163. If a > 1 thennP
k=0
�
a2k+1
�
(nk)2
2k≥ (2nn )
2
1a−1
− 2n+1
a2n+1−1
.
Mihaly Bencze
PP30164. If a1 = 3, a2 = 5 and 2an+1 =12
�a2n + 1
�for all n ≥ 1 then
compute limn→∞
n
�12 − (an+1 − 1)
nPk=1
11+ak
�.
Mihaly Bencze
PP30165. Compute limn→∞
n
�2−
nPk=2
1Fk−1Fk+1
−nP
k=1
Fk
Fk+1Fk+2
�where Fk
denote the kth Fibonacci number.
Mihaly Bencze
PP30166. If x, y ∈ R and a ∈ (0, 1) and x0 = x, y0 = y,xn+1 = axn + (1− a) yn, yn+1 = (1− a)xn + ayn for all n ∈ N thenxn = x+y
2 + x−y2 (2a− 1)n and yn = x+y
2 + y−x2 (2a− 1)n for all n ∈ N.
Mihaly Bencze
PP30167. If ak > 0 (k = 1, 2, ..., n) then determine all λ > 0 for whichQ
cyclic
�λa1
a2+a3+...+an+ 1
�≥
�λ
n−1 + 1�n
.
Mihaly Bencze
PP30168. Prove that tg
� ∞Pn=1
arctg 1F2n+1
�= 1 when Fk denote the kth
Fibonacci number.
Mihaly Bencze
Proposed Problems 1097
PP30169. If ak > 0 (k = 1, 2, ..., n) thennP
k=1
ank
1+an+1k
+
n�
k=1ak
1+n�
k=1
an+1k
≤ n+12 .
Mihaly Bencze
PP30170. If 0 < a ≤ b thenbRa
(x2+1)dx2x6+x3+3
≥ 29 ln
b3+1a3+1
.
Mihaly Bencze
PP30171. If 0 < a ≤ b then 29arctg
a3+b3
1−(ab)3≤
bRa
(x3+1)dx2x6+x3+3
.
Mihaly Bencze
PP30172. Prove that 2nP
k=1
12k+1 < ln (n+ 1) < 1
2
nPk=1
2k+1k(k+1) .
Mihaly Bencze
PP30173. Prove that 2nP
k=1
ln k(k+1)2k+1 < ln2 (n+ 1) < 1
2
nPk=1
(2k+1) ln k(k+1)k(k+1) .
Mihaly Bencze
PP30174. Prove thatnP
k=1
(2k+1)(2k6+k3+3)(k+1)2(k3+1)(k6+1)
≤ 3n(n+2)
2(n+1)2.
Mihaly Bencze
PP30175. Prove that∞Pk=1
2k6+k3+3(k3+1)(k6+1)
≤�π2
�2.
Mihaly Bencze
PP30176. Solve in Z the equation x2 + y2 + z2 + t2 = (x+ y)2(z + t)2.
Mihaly Bencze
PP30177. Solve in Z the equation x3 + y3 + z3 + t3 = (xyzt)3 .
Mihaly Bencze
1098 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP30178. If xk > 0 (k = 1, 2, ..., n) thennP
k=1
x2k(2x
6k+x3
k+3)
(x3k+1)(x6
k+1)
≤ 3n2 .
Mihaly Bencze
PP30179. Let xk > 0 (k = 1, 2, ..., n) such that limn→∞
xn = 0 and∞Pn=1
xn = +∞. Prove that exist a subsequence (xnk)k≥1 of sequence (xn)n≥1
for which∞Pk=1
x2p+1nk
= +∞ and∞Pk=1
x2p+2nk
< +∞ when p ∈ N.
Mihaly Bencze
PP30180. Prove thatP
1≤i<j≤n
1
(ni)+(nj)
≥ n2(n−1)4(2n−1) .
Mihaly Bencze
PP30181. Prove thatP
1≤i<j≤n
1
(ni)2+(nj)
2 ≥ n2(n−1)
4((2nn )−1).
Mihaly Bencze
PP30182. We consider Fn : (a,+∞) → (a,+∞) when a > 0 such thatFn (x) = a+ (x− a)a
n
. Prove that Fm ◦ Fn = Fm+n for all m,n ∈ Z. IfG = {Fn|n ∈ Z} , then (G, ◦) is abelian group and (G, ◦) ∼= (Z,+) .
Mihaly Bencze
PP30183. If xk ∈ (0, 1) (k = 1, 2, ..., n), thennP
k=1
arctgxk >π
�
n�
k=1xk
�
2
�
n+n�
k=1x2k
� .
Mihaly Bencze
PP30184. If xk ∈�0,√3�(k = 1, 2, ..., n) , then
nPk=1
arctgxkarctg1xk
≤πn2
n�
k=1
xk
2
�
n2+
�
n�
k=1
xk
�2� .
Mihaly Bencze
Proposed Problems 1099
PP30185. In all triangle ABC holdsP �
A2 + 1�arctgAarctg 1
A ≤ π2
2 .
Mihaly Bencze
PP30186. In all triangle ABC holdsP
tg�
πA2
2(A2+π2)
�< π
2 .
Mihaly Bencze
PP30187. Determine all x, y > 0 for which
arctgxarctg 1y + arctgyarctg 1
x < π2
�x
x2+1+ y
y2+1
�.
Mihaly Bencze
PP30188. Denote pk the kth prime, and M =
�x ∈ R|x =
nPk=1
1prk, r ∈ N
�.
Show that M is neither an open set nor a closed set in R with the usualtopology.
Mihaly Bencze
PP30189. Compute2πR0
sinx cos 2x sin 4xdx
(1+sin2 x)(1+sin2 2x)(1+sin2 4x).
Mihaly Bencze
PP30190. Prove that
limx→0
1x2
��ax+bx
2
� 1x −
√ab− 1
8
�ln a
b
�2√abx
�= 1
128
�ln a
b
�4√ab for all a, b > 0.
Mihaly Bencze
PP30191. Determine the best constants c1, c2 > 0 such that in all triangleABC holds 2 (4R+ r)2 (2R− r) ≥ s2 (c1R− c2r) .
Mihaly Bencze
PP30192. Prove thatnP
k=1
(k+1)2k−k2k
(k(k+1))k≥ 2 ln (n+1)n
n! .
Mihaly Bencze
1100 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP30193. If Fk and Lk denote the kth Fibonacci, respective Lucasnumbers, then
1).nP
k=1
(Fk+1−Fk)Fk+2
FkFk+1≥ 2 lnFn+1
2).nP
k=1
(Lk+1−Lk)Lk+2
LkLk+1≥ 2 lnLn+1
Mihaly Bencze
PP30194. If a ≥ 1 thennP
k=1
�ak − 1
ak
�≥ n (n+ 1) ln a.
Mihaly Bencze
PP30195. In all triangle ABC holds s2+r2+4Rr2sr ≥ s
R + 2 ln 2R2
sr .
Mihaly Bencze
PP30196. Solve in R the following system:
a41 +118 = a2
√3 +
�a23 − 1
�2+�√
2a4 −q
38
�2
a42 +118 = a3
√3 +
�a24 − 1
�2+�√
2a5 −q
38
�2
−−−−−−−−−−−−−−−−−−−a4n + 11
8 = a1√3 +
�a22 − 1
�2+�√
2a3 −q
38
�2
.
Mihaly Bencze
PP30197. If ak > 0 (k = 1, 2, ..., n) thenP
cyclic
a1�a42 − a2
√3 + 2
�≥ 5
8
nPk=1
ak.
Mihaly Bencze
PP30198. If ak > 0 (k = 1, 2, ..., n) thenP
cyclic
r(a31+a32)(a33+a34)
(a21+a1a2+a22)(a23+a3a4+a24)≥ 2
3
P√a1a3.
Mihaly Bencze
PP30199. If ak > 0 (k = 1, 2, ..., n) thennP
k=1
�a4k − ak
√3�+ 11n
8 ≥ 2nP
k=1
����a2k − 1
� �√2ak −
q38
�.���
Mihaly Bencze
Proposed Problems 1101
PP30200. If x ∈�0, π2
�then
x6+(π2−x)
6
x2(π2−x)
2�
x4+x2(π2−x)
2+(π
2−x)
4� ≥ (sinx+cosx)(1−sinx cosx)
3 sin2 x cos2 x.
Mihaly Bencze
PP30201. If x ∈�0, π2
�and λ ≥ 0 then
x sinx+ (λ+ 1)x2 cosx ≤ (λ+ 2) sin2 x.
Mihaly Bencze
PP30202. If a, b, c > 0 thenP a3+(b+c)3
a2+b2+c2+ab+2bc+ca≥ a+ b+ c.
Mihaly Bencze
PP30203. In all acute triangle ABC holds�A2 +B2 + C2
�2 ≤ (s2−r(4R+r))(4s2r2−2(s2−r2−4Rr)(s2−r2−4Rr−4R2))8R2(s2−(2R+r)2)
2 .
Mihaly Bencze
PP30204. In all triangle ABC holds�A2 +B2 + C2
�2 ≤ 2(8R2+r2−s2)((4R+r)2−2s2)s2R2 .
Mihaly Bencze
PP30205. In all acute triangle ABC holds
1).P tgA
A2 ≥�s2+r2+Rr
2sr
�2− 4R
r
2).P ctgA
(π2−A)
2 ≥�
s2+r2−4R2
s2−(2R+r)2
�2− 8R(R+r)
s2−(2R+r)2
3).P �
sinAA
�2 ≥ s2−r2−4Rr2sr 4).
P�cosAπ2−A
�2≥ 2sr
s2−(2R+r)2
Mihaly Bencze
PP30206. In all acute triangle ABC holds1).
P A2
sinA ≤ 2srs2−(2R+r)2
2).P A2
tgA ≤ sR
3).P (π
2−A)
2
cosA ≤ s2−r(4R+r)2sr 4).
P (π2−A)
2
ctgA ≤ R+rR
Mihaly Bencze
1102 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP30207. If x ∈�0, π2
�then
√sinxtgx+
√cosxctgx ≥ π
2 .
Mihaly Bencze
PP30208. Determine all x, y ∈�0, π2
�for which x2+ y2 ≤ sinxtgy+sin ytgx.
Mihaly Bencze
PP30209. If x ∈�0, π2
�then x2
�π2 − x
�2 ≤ sinx cosx.
Mihaly Bencze
PP30210. If x ∈�0, π2
�then 2x2 − πx+
�π2
�2 ≤ (sinx+cosx)(1−sinx cosx)sinx cosx .
Mihaly Bencze
PP30211. If x ≥ 0 then (arctgx)2 ≤ x2√1+x2
.
Mihaly Bencze
PP30212. In all acute triangle ABC holdsPAB ≤P sinAtgB +
PtgA sinB.
Mihaly Bencze
PP30213. If x ∈�0, π2
�then x2 sinx+ sin 2x
2 ≤ sinx+ x (1− cosx) .
Mihaly Bencze
PP30214. In all acute triangle ABC holds
1).P
A2 cosA ≤ s2−r(4R+r)2R2
2).P �
π2 −A
�2sinA ≤ 1− s2−(2R+r)2
2R2
Mihaly Bencze
PP30215. If x ∈�0, π2
�then�
π2
4 − (π − 1)x+ x2�sinx+
�x2 − x+ π
2
�cosx ≤ 2.
Mihaly Bencze
PP30216. If x ≥ 0 then xarctgx+ (arctgx)2 ≤ 2x2√1+x2
.
Mihaly Bencze
Proposed Problems 1103
PP30217. In all acute triangle ABC holds
1).P
A sinA+P
A2 cosA ≤ s2−r(4R+r)R2
2).P �
π2 −A
�cosA+
P �π2 −A
�2sinA ≤ 2− s2−(2R+r)2
R2
Mihaly Bencze
PP30218. Let be a, b, c, d > 0 determine all function f : R×R → R forwhich (af (a, b+ cf (c, d))) (bf (a, b) + df (c, d)) ≥
�a2 + b2
� �b2 + d2
�.
Mihaly Bencze
PP30219. If a, b, c > 0 then1
2a+1 + 1a+c+1 + 1
2b+1 + 1b+c+1 ≤ 2 + 1
a+b+c+1 + 12a+2b+c+1 .
Mihaly Bencze
PP30220. In all triangle ABC holdsP
�
tgA2
3�
tgA2+2tgB
2tgC
2
�
tgC2
≤ 4R+r5r .
Mihaly Bencze
PP30221. In all triangle ABC holdsP 1
(rahb+sr 3√2s)2+(rbhc)
2+(sr 3√2s)2 ≤ 1
2(sr 3√2s)2 .
Mihaly Bencze
PP30222. Let ABC be a triangle. Prove thatP
tg2A2 tgB2 ≥ 2m where m is
a positive root of the equation 2m3 +�4R+r
s
�m− r
s = 0.
Mihaly Bencze
PP30223. In all triangle ABC holds s2+(4R+r)2
s2− 18R
4R+r ≥ 2 ln (4R+r)3
27Rs2.
Mihaly Bencze
PP30224. In all acute triangle ABC holdsP A
sinA +P
A2ctgA ≤ 2sR .
Mihaly Bencze
PP30225. In all triangle ABC holds s2+r2−8Rrr2
− 18R2R−r ≥ 2 ln 2(2R−r)3
27Rr2.
Mihaly Bencze
1104 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP30226. In all triangle ABC holds
4R+r2R − 9s2
s2+(4R+r)2≤ 2 ln
16R2(s2+(4R+r)2)3
27s8.
Mihaly Bencze
PP30227. If xk ≥ 1 (k = 1, 2, ..., n) thennP
k=1
1xk
− n2
n�
k=1
xk
≤ 2 ln
�
1n
n�
k=1xk
�n
n�
k=1
xk
. If
xk ∈ (0, 1) (k = 1, 2, ..., n) , then holds the reverse inequality.
Mihaly Bencze
PP30228. In all triangle ABC holds sR − 18sr
s2+r2+4Rr≤ 2 ln
(s2+r2+4Rr)3
432s2r2R2 .
Mihaly Bencze
PP30229. Let ABCD be a tetrahedron inscribed in sphere with radius Rand denote r the radius of insphere. Prove that R− 3r ≥ 6Rr
R+3r lnR3r (A
refinement of Euler’s inequality.)
Mihaly Bencze
PP30230. In all triangle ABC holds Rr ≥ s2−3r2−4Rr
4Rr ≥�
a2c6sRr ≥ 2. (A
refinement of Eyler’s inequality.)
Mihaly Bencze
PP30231. In all acute triangle ABC holds
s2+r2−4R2
s2−(2R+r)2− 9R
R+r ≥ 2 ln 4(R+r)3
27R(s2−(2R+r)2).
Mihaly Bencze
PP30232. In all triangle ABC holds 2R−r2R − 9r2
s2+r2−8Rr≤ 2 ln
(s2+r2−8Rr)3
432R2r4.
Mihaly Bencze
PP30233. In all triangle ABC holds s2+r2+4Rr2sr − 9R
s ≥ 2 ln 2s2
27Rr .
Mihaly Bencze
Proposed Problems 1105
PP30234. In all acute triangle ABC holds
R+rR − 9(s2−(2R+r)2)
s2+r2−4R2 ≤ 2 ln(s2+r2−4R2)
3
128R2(s2−(2R+r)2).
Mihaly Bencze
PP30235. In all triangle ABC holds1). R
r ≥ ln sr
2). s2+r2+4Rr2R ≥ 1 + 2 ln 2s2
Rr
Mihaly Bencze
PP30236. In all triangle ABC holds R− 2r ≥ 4RrR+2r ln
R2r . (A refinement of
Euler’s inequality).
Mihaly Bencze
PP30237. In all acute triangle ABC holdss2+r2−4R2
s2−(2R+r)2≥ R+r
R + 2 ln 4R2
s2−(2R+r)2.
Mihaly Bencze
PP30238. In all triangle ABC holds
1). s2+r2−8Rrr2
≥ 2R−r2R + 4 ln 4R
r
2).�4R+r
s
�2 ≥ 1 + r2R + 4 ln 4R
s
Mihaly Bencze
PP30239. Prove thatnP
k=1
k(k2+1)k4+k2+1
≥ 12 ln
�n2 + n+ 1
�.
Mihaly Bencze
PP30240. Let ak > 0 (k = 1, 2, ..., n) and A = 1n
nPk=1
ak, G = n
snQ
k=1
ak,
H = nn�
k=1ak
. Prove that (A−G)(G−H)(lnA−lnG)(lnG−lnH) ≥
4AG2H(A+G)(G+H) .
Mihaly Bencze
1106 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP30241. Let ABCD be a quadrilateral inscribed in a circle with radiusR = 1, and M ∈ Int (ABCD). Prove that6 ≤ 2 (AM + CM) +BM +DM ≤ 8.
Mihaly Bencze
PP30242. Prove thatnP
k=1
2k+1k(k+1) ≥ 2 ln (n+ 1) .
Mihaly Bencze
PP30243. If A ∈ Mn (C) and Am = (−1)m In, m ∈ N,m ≥ 2 thendetermine all λk ∈ C (k = 1, 2, ..., n) for whichn ≤ rang (In + λ1A) + rang (In + λ2A) + ...+ rang (In + λnA) ≤ (m− 1)n.
Mihaly Bencze
PP30244. If a0 = 1 andnP
k=1
�nk
�an−k = 0 then study the convergence of
sequence bn = a01 + a1
1! +a22! + ...+ an
n! .
Mihaly Bencze
PP30245. Let be Hn =nP
k=1
and xn = 1H1
+ 1H2
+ ...+ 1Hn
+ ln 1Hn
. Study the
convergence of sequence (xn)n≥1 and compute its limit.
Mihaly Bencze
PP30246. If 0 < a ≤ b then 2 ln b3+1a3+1
+ arctg b3−a3
1+b3a3≤ 9
2 (b− a) .
Mihaly Bencze
PP30247. If xk > 0 (k = 1, 2, ..., n) thenP x1x2
2
x31+x3
2≤ n
2 .
Mihaly Bencze
PP30248. 1). If a, b, c, d > 0 thenQ
(a+ b+ c− d) ≤Q (a+ b)
2). If a, b, c, d > 0 then determine all λ ∈ R for whichQ(a+ b+ c+ λd) ≤Q (a+ b) .
Mihaly Bencze
Proposed Problems 1107
PP30249. Prove that1). n
�n−2p−1
�+ n2
�n−2p−2
�= p2
�np
�
2). (p− 1)n�n+1p
�+ (n+ 1)
�np
�− n2
�n
p−1
�=
�np
�
Mihaly Bencze
PP30250. Solve in N the equation (n!) + (2n)! + (3n)! = m6.
Mihaly Bencze
PP30251. If b ≥ a ≥ 67 then ln�
2 ln b−12 ln a−1
�≤
bRa
π(x)x2 dx ≤ ln
�2 ln b−32 ln a−3
�.
Mihaly Bencze
PP30252. If En (x) =nP
k=0
xk
k! then limn→∞
n�
k=0k(ex−Ek(x))
n�
k=0(ex−Ek(x))
= x2 .
Mihaly Bencze
PP30253. 1). Prove that2nPk=1
tg kπ2n+1 tg
(k+1)π2n+1 = − (2n+ 1)
2). Compute3nPk=1
tg kπ3n+1 tg
(k+1)π3n+1 tg (k+2)π
3n+1
Mihaly Bencze
PP30254. Let ABCD be a convex quadrilateral inscribed in a circle withradius R, and denote r1, r2, r3, r4 the radii of incircles of triangles ABD,BDC, ADC, ABC. Prove that AB2 +AC2 +AD2 +BC2 +BD2 + CD2 ≥≥ 2 (R+ r1)
2 + 2 (R+ r2)2 + 2 (R+ r3)
2 + 2 (R+ r4)2 .
Mihaly Bencze
PP30255. Prove that 2nP
k=1
12k+1 < ln (n+ 1) < 1
3
nPk=1
(3k2+3k+1)(2k+1)
k(k+1)(2k2+2k+1).
Mihaly Bencze
PP30256. Compute∞Pn=0
cosnx(n+1)5
, when x ∈ [0, 2π] .
Mihaly Bencze
1108 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP30257. Compute
1).∞Pn=0
1
Γ�
n+ pq
�
2).∞Pn=0
1
Γ�
n+ 1p
�
Γ�
1+ 1q
�
where p > q > 3 are given prime
Mihaly Bencze
PP30258. If λ ∈ (−∞, 0) ∪ (1,+∞) then
1 +�12
�2λ+�13
�2λ+ ...+
�1n
�2λ ≥ 1nλ−1
�π2
6 −∞Pk=0
Bk
nk+1
�λ
.
Mihaly Bencze
PP30259. In all triangle ABC holds1). (4R+ r)
�s2 + 48r2
�≥ 25s2r
2).�s2 + r2 + 4Rr
� �s2 + 24Rr2
�≥ 50s2Rr
Mihaly Bencze
PP30260. Denote pn the nth prime. Study the convergence of the sequence
xn =
��nP
k=1
1pk
��λ
− λ lnn where λ ∈ R, and {·} denote the fractional part.
Mihaly Bencze
PP30261. In all tetrahedron ABCD holds1).
Pha ≥ 791rhAhBhChD
6000r4+26hAhBhChD
20.P
rA ≥ 791rrArBrCrD750r4+52rArbrCrD
Mihaly Bencze
PP30262. Let f : [0, 1] → R be an increasing function. Prove that
n1R0
f (x) dx ≤nP
k=1
(k + 1)1R0
xkf (x) dx.
Mihaly Bencze
Proposed Problems 1109
PP30263. If x > 0 and H (x) =∞Pn=0
xn
n!(n+1)!(n+2)! , then compute
limx→∞
xλH(x)
exp(µ√x)
where λ, µ > 0.
Mihaly Bencze
PP30264. Let p ≥ 13 be a prime. Determine all k ∈ N for whichp−12P
i=1
1ik
= AB , with (A,B) = 1 result that p divides A.
Mihaly Bencze
PP30265. If a1 = 2 and an+1 = 2an +p3 (a2n − 1) then compute
∞Pn=1
1an.
Mihaly Bencze
PP30266. Compute∞R0
(sinx)2 sin�x2
�dx.
Mihaly Bencze
PP30267. Solve in Z the equation x(y+z)xn+1 + y(z+x)
yn+1 + z(x+y)zn+1 = 3.
Mihaly Bencze
PP30268. If x, a > 1 then∞Pk=0
xk
Γ(a+k) =(a−1)exx1−a(Γ(a−1)−Γ(a−1,x))
Γ(a) .
Mihaly Bencze
PP30269. Let f : [0, 1] → R be an integrable function such that1R0
f (x) dx = 0. Determine the best constants c1, c2 ≥ 0 such that
c1
�1R0
xf (x) dx
�2
≤1R0
f2 (x) dx ≤ c2
�1R0
xf (x) dx
�2
.
Mihaly Bencze
PP30270. If a > 1 thennQ
k=2
ln ak+1a+1
ln ak+1+12
≥ 2n(n+1) .
Mihaly Bencze
1110 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP30271. Let be p ≥ 3 a prime. The numbers n, n+ 1, n+ 2 are p-healty ifand only if
�(n+ 2)
√p�+�n√p�− 2
�(n+ 1)
√p�= 1 where [·] denote the
integer part. What is the probability that three consecutive numbers to bep-healty?
Mihaly Bencze
PP30272. If n ≥ 3 then the equation xn + 2xn−1 + 3xn−2 + ...+ nx− 1 = 0
have a root xn ∈ (0, 1) . Compute∞Pn=1
1x2n.
Mihaly Bencze
PP30273. If ak ≥ 80, (k = 1, 2, ..., n) thennQ
k=1
π (a1a2...ak) ≥ (π (a1))n (π (a2))
n−1 ... (π (an−1))2 π (an) .
Mihaly Bencze
PP30274. In all triangle ABC holdstg πA
π2+A2 + tg πBπ2+2(A2+B2)
+ tg πCπ2+3(A2+B2+C2)
< 3.
Mihaly Bencze
PP30275. In all triangle ABC holdstg 2as
a2+4s2+ tg 2bs
42+2(a2+b2)+ tg cs
5s2−3r2−12Rr< 3.
Mihaly Bencze
PP30276. In all triangle ABC holds
1). tg rrar2+r2a
+ tg rr2arbr2ar
2b+2r2(r2a+r2
b)+ tg s2r
2rc(2s2−12Rr−3r2)< 3
2). tg rha
r2+h2a+ tg rh2
ahb
h2ah
2b+2r2(h2
a+h2b)
+ tg s2rhc(5s2−3r2−12Rr)
< 3
Mihaly Bencze
PP30277. In all triangle ABC holds
tg
�2R−r2R
sin2 A2
( 2R−r2R )
2+(sin A
2 )4
�+ tg
2R−r2R
sin2 B2
( 2R−r2R )
2+2
�
(sin A2 )
4+(sin B
2 )4�
!+
+tg
�2R(2R−r) sin2 C
232R2−8Rr+5r2−3s2
�< 3.
Mihaly Bencze
Proposed Problems 1111
PP30278. In all triangle ABC holdstg s(s−a)
s2+(s−a)2+ tg s(s−b)
s2+2((s−a)2+(s−b)2)+ tg s(s−c)
4s2−6r2−24Rr< 3.
Mihaly Bencze
PP30279. In all triangle ABC holdstg (4R+r)ra
(4R+r)2+r2a+ tg (4R+r)rb
(4R+r)2+2(r2a+r2b)
+ tg (4R+r)rc4(4R+r)2−6s2
< 3.
Mihaly Bencze
PP30280. If xk > 0 (k = 1, 2, ..., n) then tg x1(x1+x2+...+xn)
(x1+x2+...+xn)2+x2
1
+
+tg x2(x1+x2+...+xn)
(x1+x2+...+xn)2+2(x2
1+x22)
+ ...+ tg xn(x1+x2+...+xn)
(x1+x2+...+xn)2+n(x2
1+x22+...+x2
n)< n.
Mihaly Bencze
PP30281. In all triangle ABC holds tg
�4R+r2R
cos2 A2
( 4R+r2R )
2+(cos A
2 )4
�+
+tg
4R+r2R
cos2 A2
( 4R+r2R )
2+2
�
(cos A2 )
4+(cos B
2 )4�
!+ tg
�4R(4R+r) cos2 C
2
5(4R+r)2−3s2
�< 3.
Mihaly Bencze
PP30282. In all triangle ABC holds
1). 2R−r2R + 3r2
s2+r2−8Rr≥ 4 3
q�r4R
�2
2). 4R+r2R + 3s2
s2+(4R+r)2≥ 3 3
q�s4R
�2
Mihaly Bencze
PP30283. If xk > 0 (k = 1, 2, ..., n) thennP
k=1
xk +n
n�
k=1
1xk
≥ (n+ 1) n
snQ
k=1
xk.
Mihaly Bencze
PP30284. In all triangle ABC holds1). s+ 6sRr
s2+r2+4Rr≥ 2 3
√4sRr
2). s(R+r)4R+r ≥ 3
√sr2
3). R+ r ≥ 3√s2r
Mihaly Bencze
1112 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP30285. Prove thatnP
k=1
1k+1
�e1+
12+...+ 1
k−c − k − 1
2
�< n
24(n+1) <nP
k=1
1k
�e1+
12+...+ 1
k−c − k − 1
2
�
when C is the Euler constant.
Mihaly Bencze
PP30286. Compute limn→∞
n
ln 2−
2n−1Pk=1
12n−1+k
!.
Mihaly Bencze
PP30287. Prove thatmP
n=1
nPk=0
(nk)2
n(2n−1k )
2 ≥ 4mm+1 .
Mihaly Bencze
PP30288. In all acute triangle ABC holds 2 <P �
π2 −A
�tgA < π2
2 .
Mihaly Bencze
PP30289. In all acute triangle ABC holdsP 2
π(π−A)2+π
2A2
A sinA ≤ 4π ≤P (π−A)2+A2
A sinA.
Mihaly Bencze
PP30290. In all acute triangle ABC holdsP 2
π(π−A)2−π
2A2
2π(π−A)2+π
2A2
≤ 1 + rR ≤P (π−A)2−A2
(π−A)2+A2.
Mihaly Bencze
PP30291. In all acute triangle ABC holds π ≤P (π −A) tgA2 ≤ π2
2 .
Mihaly Bencze
PP30292. In all triangle ABC holdsP(1 + sinA) arctg (sinA) ≤ πs
2R ≤P�π2 + sinA
�arctg (sinA) .
Mihaly Bencze
PP30293. In all acute triangle ABC holdsP(1 + cosA) arctg (cosA) ≤ π(R+r)
2R ≤P �
π2 + cosA
�arctg (cosA) .
Mihaly Bencze
Proposed Problems 1113
PP30294. In all acute triangle ABC holds ln π−Aπ−B ≤
BRA
tg x2dx
x ≤ π2 ln
π−Aπ−B and
his permutations.
Mihaly Bencze
PP30295. Prove thatπ2
nPk=1
1�
π2+ 1
k(k+1)
�
arctg 1k(k+1)
≤ nn+1 ≤ π
2
nPk=1
1�
1+ 1k(k+1)
�
arctg 1k(k+1)
.
Mihaly Bencze
PP30296. If a1 = 1 and an+1 = 1 + 12
�1 + 1
n+1
�an for all n ≥ 1 then study
the convergence of the sequence bn = a1(n1)
+ a2(n2)
+ ...+ an(nn)
.
Mihaly Bencze
PP30297. If a1 = a2 = 1, a3 = 4 and an+3 = 2an+2 + 2an+1 − an for alln ≥ 1 then compute
1).nP
k=1
a3k 2).
n∞Pk=1
1ak
3).∞Pk=1
1a2k
Mihaly Bencze
PP30298. If a, b > 0 then determine all x ∈ R for which(1 + a2 − x)(1 + b2 − x) ≤ (1 + ab)2.
Mihaly Bencze
PP30299. If x, y ∈ (0, π2 ) then:
(sin2 x+sin2 y)sin2 + sin2 y ·(cos2 x+cos2 y)cos
2 x+cos2 y
(sinx)2 sin2 x·(sin y)2 sin2 y ·(cosx)2 cos2 x·(cos y)2 cos2 y≤ 4
Daniel Sitaru
PP30300. If a, b, c > 0; a+ b+ c = 3; 0 ≤ x ≤ 1 then:a�ba
�x+ b
�cb
�x+ c
�ac
�x+ b
�ab
�x+ c
�bc
�x+ a
�ca
�x ≤ 6
Daniel Sitaru
PP30301. Find: Ω =P∞
k=1
�P∞n=1n6=k
�1
n2−k2
��
Daniel Sitaru
1114 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP30302. Find: Ω =R �P∞
n=1
�3n sinh3
�x3n
���dx
Daniel Sitaru
PP30303. If x, y, z, t > 0 then:
(xy + yz + zt+ tx)�
1x4 + 1
y4+ 1
z4+ 1
t4
�≥
�1x + 1
y + 1z + 1
t
�2
Daniel Sitaru
PP30304. If a, b, c, d > 0 then:(ab+ bc+ cd+ da)(a4 + b4 + c4 + d4) ≥ abcd(a+ b+ c+ d)2
Daniel Sitaru
PP30305. Let be: Ω(a, b, c) =P∞
n=1an2+bn+c
n! ; a, b, c > 0
Prove that: Ω(a, b, c) + Ω(b, c, a) + Ω(c, a, b) ≥ 3(4e− 1) 3√abc
Daniel Sitaru
PP30306. In ΔABC the following relationship holds:P
CyC(a,b,c) aq�
(b− c)2 + 4r2��(c− a)2 + 4r2
�≥ abc
Daniel Sitaru
PP30307. In acute ΔABC the following relationship holds:2sinA + ssinB + 2sinC + 2cosA + 2cosB + 2cosC > 9
Daniel Sitaru
PP30308. If a1, a2, ..., a8 ≥ 1 then: a41 + a42 + ...+ a48 ≤ (a1a2...a8)4 + 7
Daniel Sitaru
PP30309. Find: Ω = limn→∞ npΩn(a)− (a+ 1)!; a ∈ N ;
Ωn(a) =Pn
k=0(k2 − a2 + 1)(a+ k)!; n ∈ N
Daniel Sitaru
PP30310. Let be Ω(a) = limn→∞Pn
k=1
�1
3k sin3(3k sin a)
�Prove that if
a, b, c ∈ [0, π2 ] then: 4�bΩ(a) + cΩ(b) + aΩ(c)
�≤ 3(a2 + b2 + c2)
Daniel Sitaru
Proposed Problems 1115
PP30311. If 0 < a ≤ b ≤ c then: 1(a−b+c)6
+ 1b6
≤ 1a6
+ 1c6
Daniel Sitaru
PP30312. If a, b > 0, a2 + b2 = 1 then: 1a + 1
b ≥ 2√2
Daniel Sitaru
PP30313. If x, y > 0; Ω(x, y) =P∞
n=12n2+(2x+2y+5)n+2xy+6x−y3n(n+y)(n+y+1)(n+y+2) then:
Ω(y, x) ≤ 19 3√xy
Daniel Sitaru
PP30314. If a, b, c, d > 0; a3 + b3 + d3 = 1 then: a+b+c+dabcd ≥ 16
Daniel Sitaru
PP30315. If a, b, c, > 0; a3 + b3 + c3 = 1 then: a+b+cabc ≥ 3 3
√9
Daniel Sitaru
PP30316. If a, b, c, d > 0; a2 + b2 + c2 + d2 = 1 then: a+b+c+dabcd ≥ 32
Daniel Sitaru
PP30317. If a, b, c > 0; a2 + b2 + c2 = 1 then: a+b+cabc ≥ 9
Daniel Sitaru
PP30318. If a, b, c > 0; x, y, z, t ∈ R then:(a+b+c)y2
a + (a+b+c)z2
b + (a+b+c)t2
c ≥ x(2y + 2z + 2t− x)
Daniel Sitaru
PP30319. If a, b > 0; x, y, z ∈ R then: (a+b)y2
a + (a+b)z2
b ≥ x(2y + 2z − x)
Daniel Sitaru
PP30320. In ΔABC the following relationship holds:3
qsinAsinB + 3
qsinBsinC + 3
qsinCsinA − 3
qsinAsinC − 3
qsinBsinA − 3
qsinCsinB < 1
Daniel Sitaru
1116 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP30321. If ai > 0 (i = 1, 2, ..., n) and λ1,λ2, ...,λk > 0 such thatλ1 + λ2 + ...+ λk = 1 then
P λ1a2+λ2a3+...+λkak+1
a21≥
nPi=1
1ai
≥P 1λ1a1+λ2a2+...+λkak
.
Mihaly Bencze
PP30322. In all acute triangle ABC holds
P 1a2(b2+c2−a2)
≥ (s2+r2+4Rr)2
32s2R2r2(s2−r2−4Rr).
Mihaly Bencze
PP30323. Let be a1 = 2 and (n+ 2) an+1 = nan +�n2 + 2n+ 2
�n! for all
n ≥ 1. Compute∞Pk=1
1ak.
Mihaly Bencze
PP30324. In all acute triangle ABC holdsP 1
sin 2A ≥ s2+r2+4Rr2sr .
Mihaly Bencze
PP30325. In all triangle ABC holds
1).Q �
ctgA2
�a2 ≤�
2s(4R+r)s2−r2−4Rr
�2(s2−r2−4Rr)
2).Q
actgA2 ≤ (4r (4R+ r))
s2r
Mihaly Bencze
PP30326. In all triangle ABC holds
1).Q �
tgA2
�a3 ≤�s2(2R−3r)+r2(4R+r)
s2−3r2−6Rr
�2s(s2−3r2−6Rr)
2).Q
atgA2 ≤
�2s(s2(2R−3r)+r2(4R+r))
4R+r
� 4R+r3s
Mihaly Bencze
Proposed Problems 1117
PP30327. In all triangle ABC holds
1).Q �
ctgA2
�a3 ≤�s2(2R+3r)−r(4R+r)2
s(s2−3r2−6Rr)
�2s(s2−3r2−6Rr)
2).Q
actgA2 ≤
�2r(s2(2R+3r)−r(4R+r)2)
s
� s3r
Mihaly Bencze
PP30328. In all triangle ABC holds
1).P
ctgA2 ≥ s
R−r 2).P
tgA2 ≥ s
4R+r
Mihaly Bencze
PP30329. Solve in Z the equation 1x(y+1) +
1(y+2)(z+3) =
1996 .
Mihaly Bencze
PP30330. In all triangle ABC holds
1).Q �
tgA2
�a2 ≤�
2s(R−r)s2−r2−4Rr
�2(s2−r2−4Rr)
2).Q
atgA2 ≤
�4s2(R−r)4R+r
� 4R+r2s
Mihaly Bencze
PP30331. In all triangle ABC holdsP√
a ≤ min
�q2s(2R−r)
r ; (4R+ r)q
2s
�.
Mihaly Bencze
PP30332. In all triangle ABC holds
1).Q �
tgA2
�a ≤�2R−r
s
�2s2).
Qatg
A2 ≤
�2s(2R−r)4R+r
� 4R+r2s
Mihaly Bencze
PP30333. In all triangle ABC holds
1).Q �
ctgA2
�a ≤�4R+r
s
�2s
2).Q
actgA2 ≤
�2r(4R+r)
s
� sr
Mihaly Bencze
1118 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP30334. In all triangle ABC holds
1).P 1
(a+1)(b+1)(a+b+1) ≤5√5−112Rr
2).P 1
(ra+1)(rb+1)(ra+rb+1) ≤(5
√5−11)(4R+r)
2s2r
3).P 1
(1+sin2 A2 )(1+sin2 B
2 )(1+sin2 A2+sin2 B
2 )≤ 4R(2R−r)(5
√5−11)
r2
4).P 1
(1+cos2 A2 )(1+cos2 B
2 )(1+cos2 A2+cos2 B
2 )≤ 4R(4R+r)(5
√5−11)
s2
Mihaly Bencze
PP30335. In all triangle ABC holds
1). 4R+r2 +
P r2arb+rc
≥r3�(4R+ r)2 − 2s2
�
2). 12 + r
P rarb(ra+rb)r2c
≥ 1s
p3 (s2 − 2r (4R+ r))
Mihaly Bencze
PP30336. If xi > 0 (i = 1, 2, ..., n) and k, p ∈ N∗ then
Pcyclic
xk1
x2+x3+...+xn+ 1
n−1
�nP
i=1xi
� 1p
≥ p+1n−1
p+1
snpp
nPi=1
xki .
Mihaly Bencze
PP30337. In all triangle ABC holds12
Pma +
P m2a
mb+mc≥ 3
2
p2 (s2 − r2 − 4Rr).
Mihaly Bencze
PP30338. In all triangle ABC holdsP cos2 A
2
cos π−A4
cos B−C2
+P
cos A2 ≥
q6�4 + r
R
�.
Mihaly Bencze
PP30339. If ak > 0 (k = 1, 2, ..., n) thenP (5a1+a2)(a1+a2)
8a1+a2≥ 4
3
nPk=1
ak.
Mihaly Bencze and Sitaru Daniel
PP30340. If ak > 0 (k = 1, 2, ..., n) thenP (5a1+a2)(8a1+a2)
a1+a2≥ 27
nPk=1
ak.
Mihaly Bencze
Proposed Problems 1119
PP30341. If ak > 0 (k = 1, 2, ..., n) andnQ
k=1
ak = 1 then
nPk=1
ak(1+a1)(1+a2)...(1+ak)
≥ 2n−12n .
Mihaly Bencze and Sitaru Daniel
PP30342. In triangle ABC, D ∈ (BC) , E ∈ (CA) , F ∈ (AB) andAD,BE,CF are Gergonne’s cevians. Prove that
r (4R+ r) +P
AD2 ≥P�b2(s−b)c2(s−c)
� 1a .
Mihaly Bencze
PP30343. If xi > 0 (i = 1, 2, ..., n) and k ∈ N, k ≥ 2 then
(k + 1)P
cyclic
√x2
x1+kx2≤ P
cyclic
��x21 − x1x2 + x22
�x2k+11 xk
2−22
� −1
2(k+1)2
Mihaly Bencze and Sitaru Daniel
PP30344. In triangle ABC, D ∈ (BC) , E ∈ (CA) , F ∈ (AB) andAD,BE,CF are the Gergonne’s cevians. Prove thatP
a ·AD2 = 2sr (2R+ 5r) .
Mihaly Bencze
PP30345. In all triangle ABC holdsP ma
(b+c)sa≥ s2+r2+4Rr
8sRr .
Mihaly Bencze
PP30346. In all triangle ABC holdsP (b+c)sa
ma≤ (s2+r2+4Rr)
2+8s2Rr
8s(s2+r2+2Rr).
Mihaly Bencze
PP30347. Prove thatnP
k=1
sin
1
k2(k+1)2−1R0
e−t2dt
≤ n
n+1 .
Mihaly Bencze
1120 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP30348. Let ABC be an acute triangle and M ∈ Int (ABC) . DenoteRa, Rb, Rc the circumradii of triangles BMC, CMA, AMB. Prove thatM ≡ H if and only if Ra = Rb = Rc where H is the orthocentrum of triangleABC.
Mihaly Bencze
PP30349. Prove thatnP
k=1
cos
√
(k2+k)2−1R0
e−t2dt
≥ n
n+1 .
Mihaly Bencze
PP30350. If 0 < a ≤ b thenbRasin
�xR0
e−t2dt
�dx ≤ (b−a)(b+a)√
1+a2+√1+b2
.
Mihaly Bencze
PP30351. If 0 < a ≤ b thenbRax cos
�xR0
e−t2dt
�dx ≥ (b−a)(b+a)√
1+a2+√1+b2
.
Mihaly Bencze
PP30352. If x ∈�0, π2
�then 2 sin
tgxR0
e−t2dt
!sin
ctgxR0
e−t2dt
!≤ sin 2x.
Mihaly Bencze
PP30353. If x ∈�0, π2
�then 2 cos
tgxR0
e−t2dt
!cos
ctgxR0
e−t2dt
!≥ sin 2x.
Mihaly Bencze
PP30354. If x ∈�0, π2
�then cos2
tgxR0
e−t2dt
!+ cos2
ctgxR0
e−t2dt
!≥ 1.
Mihaly Bencze
PP30355. If x ∈�0, π2
�then sin2
tgxR0
e−t2dt
!+ sin2
ctgxR0
e−t2dt
!≤ 1.
Mihaly Bencze
Proposed Problems 1121
PP30356. If x ∈ R thenp1 + sin4 x sin
sin2 xR0
e−t2dt
!+
√1 + cos4 x sin
cos2 xR0
e−t2dt
!≤ 1.
Mihaly Bencze
PP30357. If ak > 0 (k = 1, 2, ..., n) then
nPk=1
sin
�akR0
e−x2dx
�≤
nn�
k=1ak
�
n2+
�
n�
k=1ak
�2
.
Mihaly Bencze
PP30358. If ak > 0 (k = 1, 2, ..., n) thennP
k=1
cos
�akR0
e−x2dx
�≤ n2
�
n2+
�
n�
k=1ak
�2
.
Mihaly Bencze
PP30359. If ak > 0 (k = 1, 2, ..., n) thenP
cyclic
a1a2R0
e−x2dx ≤ π
2
nPk=1
ak
Mihaly Bencze and Sitaru Daniel
PP30360. Prove that
1).nP
k=1
tg
F 2kR
0
e−x2dx
!≤ FnFn+1
2).nP
k=1
tg
L2kR
0
e−x2dx
!≤ LnLn+1 − 2,
where Fk and Lk denote the kth Fibonacci respective Lucas numbers.
Mihaly Bencze
PP30361. If x > 0 then sin
�xR0
e−t2dt
�≤ x√
1+x2≤ x cos
�xR0
e−t2dt
�.
Mihaly Bencze
1122 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP30362. Prove thatnP
k=1
tg
1k(k+1)R0
e−x2dx
≤ n
n+1 .
Mihaly Bencze
PP30363. Prove that
1).nP
k=1
tg
FkR0
e−x2dx
!≤ Fn+2 − 1
2).nP
k=1
tg
LkR0
e−x2dx
!≤ Ln+2 − 3,
where Fk and Lk denote the kth Fibonacci respective Lucas numbers.
Mihaly Bencze
PP30364. In all tetrahedron ABCD holds
1).P 1√
ra
1√rbR0
e−x2dx ≤ π
q2r
1).P 1√
ha
1√hbR0
e−x2dx ≤ π√
r
Mihaly Bencze
PP30365. If ak > 0 (k = 1, 2, ..., n) then
Pcyclic
�a1R0
e−x2dx
�arctga2 ≤
nPk=1
arctg2ak.
Mihaly Bencze
PP30366. If 0 < a ≤ b thenbRa
�xR0
e−t2dt
�dx ≤ barctgb− aarctga− 1
2 ln1+b2
1+a2.
Mihaly Bencze
PP30367. In all triangle ABC holds
1).P√
a
√bR
0
e−x2dx < π
2
√6s
2).P√
ra
√rbR0
e−x2dx ≤ π
2
p3 (4R+ r)
Proposed Problems 1123
3).P√
A
√BR
0
ex2dx < π
√3π2
Mihaly Bencze and Sitaru Daniel
PP30368. If x ∈ R then(1+2 cos2 x) cos4 x
1+sin2 x+
(1+2 sin2 x) sin4 x1+cos2 x
≤ 8 + 12 sin2 x cos2 x.
Mihaly Bencze and Marius Dragan
PP30369. Prove that
√3R
1√3
�xR0
e−t2dt
�dx ≤ 5π
6√3− 1
2 ln 3.
Mihaly Bencze
PP30370. Let ABCD be a tetrahedron and M ∈ Int (ABCD) . DenoteRa, Rb, Rc the radii of circumspheres of tetrahedron BCDM, CDAM, DABM,ABCM. Prove that M ≡ H if and only if Ra = Rb = Rc = Rd where H is theorthocentrum of tetrahedron ABCD.
Mihaly Bencze
PP30371. Let ABC be an acute triangle and H denote the orthocentrum.Prove that the circumradii of triangles AHB, BHC, CHA are equal with Rthe circumradii of triangle ABC.
Mihaly Bencze
PP30372. Find all n ∈ N for which 20162n−1 + 20172n + 20182n+1 is aperfect square.
Mihaly Bencze
PP30373. Let ABCD be a convex quadrilateral and G1, G2 the centroid oftriangles ABD and BCD. The parallel lines through a point M situated inthe plane of the triangle to the medians AA1, BB1, DD1
(A1 ∈ BD,B1 ∈ AD,D1 ∈ AB) and BB2, CC1, DD2
(B2 ∈ CD, C1 ∈ BD, D2 ∈ BC) intersect lines BD,DA,AB respectiveCD,DB,BC in points A2, B2, D2 respective B3, C2, D3. Prove thatA1A
22 +B1B
22 +D1D
22 +B1B
23 + C1C
22 +D1D
23 ≥ 3
4MG21 +
34MG2
2.
Mihaly Bencze
1124 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP30374. Let ABC be a triangle such that
cos Aλ + cos B
λ + cos Cλ =
√λ2+1−1
λ . Determine all λ > 0 for which
max {A,B,C} > λ2πλ2+1
.
Mihaly Bencze
PP30375. Solve in Z the equation4�x2 + y2
�+ 4034 (x+ y)− 12xy − 4068285 = 0.
Mihaly Bencze
PP30376. Solve in Z the equationx+ y + z + (x− y)2 + (y − z)2 + (z − x)2 = xy + yz + zx.
Mihaly Bencze
PP30377. Let x be a real number such that xm�xn + xn−1 + ...+ x+ 1
�
and xn�xm + xm−1 + ...+ x+ 1
�are rational for some relatively prime
positive integers m and n. Prove that x is irrational.
Mihaly Bencze
PP30378. Let ABCD be a cyclic quadrilateral. Points E and F lie on thesides AB and BC, respectively, such that BFE∡ = λBDE. Determine allλ > 0 for which EF ·AE ·AD = AE · FC ·AD +AE2 · CD.
Mihaly Bencze
PP30379. In all triangle ABC holds 27π3 ≤ 1
ABC ≤ 27R2π3r
. (A refinement ofEuler’s inequality.)
Mihaly Bencze
PP30380. If xk ∈�0, π4
�(k = 1, 2, ..., n) then
nPk=1
ctg2xk ≥1−
n�
k=1cos2 xk
1−n�
k=1sin2 xk
.
Mihaly Bencze
PP30381. ComputeP
n,k≥0
2n+k
(22n+1)�
22k+1
� .
Mihaly Bencze
Proposed Problems 1125
PP30382. Prove that∞Pn=1
7776n5−14256n4+9072n3−2412n2+216n−1n(6n−1)! = 6
�1− 1
e
�.
Mihaly Bencze
PP30383. Prove that the sequence FF
F11
1 + 1, FF
F22
2 + 1, ..., FFFnn
n + 1, ... andan arbitrary infinite increasing arithmetic sequence have aither infinitelymany terms in common or at most one term is common. Same question if wesubstitute Fn with Ln, when Fn and Ln denote the nth Fibonacci respectiveLucas numbers.
Mihaly Bencze
PP30384. Let 0 = a0 < a1 < ... < an ≤ an+1 such thatnP
k=1
aλk = 1 when
λ ≥ 1. Prove thata2λ−11
a2−a0+
a2λ−12
a3−a1+ ...+ a2λ−1
n
an+1−an−1≥ 1
anan+1.
Mihaly Bencze
PP30385. Let 0 = a0 < a1 < ... < an < an+1 = 1 such thatnP
k=1
aλk = 1 where
λ ≥ 1. Prove thatna2λ−1
1a1−a0
+(n−1)a2λ−1
2a3−a1
+ ...+2a2λ−1
n−1
an−an−2+ a2λ−1
n
an+1−an−1≥ n1+ 1
λ .
Mihaly Bencze
PP30386. If ak,λk > 0 (k = 1, 2, ..., n) , thenpλ1a21 + λ2a22 + ...+ λna2n+
+pλ1a22 + λ2a23 + ...+ λna21 + ...+
qλ1a2n + λ2a21 + ...+ λna2n−1 ≥
≥s
nPk=1
λk
�nP
k=1
ak
�(A generalization of problem J326 Mathematical
Reflections).
Mihaly Bencze
PP30387. If ak,λk > 0 (k = 1, 2, ..., n) and r ≥ 1 then
(λ1ar1 + λ2a
r2 + ...+ λna
rn)
1r + (λ1a
r2 + λ2a
r3 + ...+ λna
r1)
1r + ...
+�λ1a
rn + λ2a
r1 + ...+ λna
rn−1
� 1r ≥
�nP
k=1
λk
� 1r�
nPk=1
ak
�.
Mihaly Bencze
1126 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP30388. If ak > 0 (k = 1, 2, ..., n) andnP
k=1
ak = n, then
Pcyclic
n−1
qan−11 + a2a3...an−1 ≤ n · 2
1n−1 .
Mihaly Bencze
PP30389. If a0 = 1 and an+1
�n2an + a2n + 1
�= an for all n ≥ 1 then
compute limn→∞
n�3− n3an
�.
Mihaly Bencze
PP30390. If a0 = 1 and an = an+1
�nkan + pa2n + 1
�for all n ≥ 1, where
k ∈ N∗, k ≥ 2 then limn→∞
nk+1an = k + 1.
Mihaly Bencze
PP30391. Solve in N the equation 1n
nPk=1
xk + n
snQ
k=1
xk = 2n.
Mihaly Bencze
PP30392. If a, b, c > 0 thenP bc
a+√a2+bc
≤ 12
P a2+bca+c .
Mihaly Bencze
PP30393. If ak > 0 (k = 1, 2, ..., n) thenP
n−1
qan−11 + a2...an ≤ 1
n−1
P an−11 +a2...ana2+...+an
+ n− 3 +nP
k=1
ak.
Mihaly Bencze
PP30394. If ak > 0 (k = 1, 2, ..., n) thenP
n−1
qan−11 + a2...an ≤ 1
n−1
P an−11 +a2...an
(a2+...+an)n−2 + (n− 2)
nPk=1
ak.
Mihaly Bencze
PP30395. If a, b, c, d > 0 and abcd+ (P
a)�P
a3�= 5 (
Pa)4 then
(P
abc) (P
a) ≤ abcd+ 3 (P
a)4 .
Mihaly Bencze
Proposed Problems 1127
PP30396. If ak > 0 (k = 1, 2, ..., n) thenP
n−1
qan−11 + a2a3...an ≤ n
n−1
nPk=1
ak.
Mihaly Bencze
PP30397. If a, b, c, d > 0 thenP
a12 + 4 (abcd)3 ≥ 2P
(abc)4 .
Mihaly Bencze
PP30398. In all triangle ABC holdsPr�
1ra
+ 1hb
�2+�
1rb
+ 1hc
��1rc
+ 1ha
�≤ 3
r .
Mihaly Bencze
PP30399. If 0 = a0 < a1 < ... < an+1 such thata1a2 + a2a3 + ...+ an−1an = anan+1 thenn−1a23−a20
+ n−2a2n−a21
+ ...+ 2a2n−a2n−3
+ 1a2n+1−a2n−2
≥ (n−1)2
a21+a22+...+a2n−1.
Mihaly Bencze
PP30400. If xk > 0 (k = 1, 2, ..., n) then snP
k=1
x2k
! P
cyclic
qx21 + x22 + ...+ x2n−1
!+ 2
�1−
qn
n−1
� P1≤i<j≤n
xixj ≥
≥ (n− 1)nP
k=1
x2k.
Mihaly Bencze
PP30401. Each of the diagonals AD, BE, CF of the convex equiangularhexagon ABCDEF divides its area in half. Prove that�AB2 + CD2 + EF 2
� �AC2 + CE2 + FA2
�=
=�BC2 +DE2 + FA2
� �BD2 +DF 2 + FB2
�.
Mihaly Bencze
PP30402. Let ABC be a convex quadrilateral in which A∡ ≥ 60◦,D∡ ≥ 60◦ and E ∈ AB,F ∈ AC,M ∈ BD,N ∈ DC. Prove that10−8 cos A+D
2cos A−D
2BD ≤ 1
min{BE,EF,FC} + 1min{BM,MN,NC} .
Mihaly Bencze
1128 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP30403. If a, b > 0 then�a+ 1
2
�2+�b+ 1
2
�2+ 23
2 ≥ 4 4
q4 (a+ b+ c)3.
Mihaly Bencze and Marius Dragan
PP30404. If a, b > 0 then (a+ 1)3 + (b+ 1)3 + 16 ≥ 4q(a+ b+ 2)3.
Mihaly Bencze and Marius Dragan
PP30405. If x, y ∈ R then
1). ch4x+ ch4y + 48 ≥ 8 4
q8 (ch2x+ ch2y)3
2). ch6x+ ch6y + 16 ≥ 4
q(ch2x+ ch2y)3
Mihaly Bencze and Marius Dragan
PP30406. Let ABCD be a convex quadrilateral such that A∡ = D∡ = 90◦.Let M,N be two points on segment AB. Lines through M and N tangent tothe circumcircle of triangles ADC and DEC intersect lines AB and BD inpoints M1,M2 respective N1, N2. Prove that AM1 + CN1 = BM2 +DN2.
Mihaly Bencze
PP30407. If x, y, z > 0 and x+ y > z, y + z > x, z + x > y then
min
�P (2z2+xy)2
2y2+2z2−x2 ;P (2z2+xy)
2
2z2+2x2−y2
�≥ 3
Px2.
Mihaly Bencze
PP30408. If ak > 0 (k = 1, 2, ..., n) then
Pcyclic
3
q2 (a1 + 3) (a2 + 3)
�a21 + 3
� �a22 + 3
�≥ 4
nPk=1
ak + 4n.
Mihaly Bencze
PP30409. Let be xk ∈ N (xi 6= xj) (i 6= j) (k = 1, 2, ..., n), n ≥ 5 andM = {xixjxkxr|1 ≤ i, j, k, r ≤ n} . Determine cardM.
Mihaly Bencze
Proposed Problems 1129
PP30410. In all triangle ABC holds
1).P (8r2a+39)(8r2b+39)
(5r4a+34r2a+80)(5r4b+34r2b+80)
≤ 4R+r4s2r
2).P (8h2
a+39)(8h2b+39)
(5h4a+34h2
a+80)(5h4b+34h2
b+80)
≤ s2+r2+4Rr16s2r2
Mihaly Bencze
PP30411. In all triangle ABC holds1).
P (3r−ra)rbrn+1a (3r+rb)
≥ 0
2).P (3r−ha)hb
hn+1a (3r+hb)
≥ 0 for all n ∈ N
Mihaly Bencze
PP30412. Prove that exist infinitely many A ∈ M3 (Q) for which Ap = I3where p ≥ 5 is a prime.
Mihaly Bencze
PP30413. Solve in R the following inequation
8x2+395y4+34z2+80
+ 8y2+395z4+34x2+80
+ 8z2+395x4+34y2+80
≤ 12
�1x + 1
y + 1z
�.
Mihaly Bencze
PP30414. Determine all prime p such that (p+ 2)5 + (p+ 1)5 − 2p5 areperfect square.
Mihaly Bencze
PP30415. Prove thatbRa
arctg(3√3tg2x)
x2+1dx ≤ ln b2+1
a2+1for all 0 < a < b < π
4 .
Mihaly Bencze
PP30416. If x, y ∈�0, π4
�then�√
tg2x+√tg2y
�cosx cos y ≥
p3√3 sin (x+ y) .
Mihaly Bencze and Marius Dragan
1130 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP30417. Determine all polynomials P ∈ R [x] for which1R0
(P (xm))1n dx =
1R0
(P (xm))1m dx, where m,n ∈ N∗
Mihaly Bencze
PP30418. Prove thatnP
k=1
tg(2(k+1)π)Rtg(2kπ)
xdx
(1+x2)32 arctgx
< n2(n+1)π .
Mihaly Bencze
PP30419. If x, y ∈�0, π4
�then
2�x2 + y2
�≥ xarctg
�3√3tg2y
�+ yarctg
�3√3tg2x
�.
Mihaly Bencze
PP30420. In all triangle ABC holdsP 4R+r−rc
4R+r+2ra+rb≤ 1.
Mihaly Bencze
PP30421. Prove that 4(m−1)4m+1 +
mPn=2
1
nn�
k=2
k√
2k+1≤ m−1
2m +mP
n=2
1
(4n+1)n�
k=2
k√
2k+1
for all m ≥ 2,m ∈ N.
Mihaly Bencze
PP30422. Solve in R the following system:
�x31 − 1
�4= x2 − 1�
x32 − 1�4
= x3 − 1−−−−−−−−�x3n − 1
�4= x1 − 1
.
Mihaly Bencze
PP30423. Solve in R the following system:
1− sin 2x1 = 16�2−
√3�(cosx2)
6
1− sin 2x2 = 16�2−
√3�(cosx3)
6
−−−−−−−−−−−−−−−1− sin 2xn = 16
�2−
√3�(cosx1)
6
.
Mihaly Bencze
Proposed Problems 1131
PP30424. In all tetrahedron ABCD holds 216√2R3 ≤
�P 1rA
��P 1r2A
�≤ 4
√2
r3.
Mihaly Bencze
PP30425. If x ∈�0, π2
�then�
x sinx+ (1 + sinx)�π2 − x
�� ��π2 − x
�cosx+ (1 + cosx)x
�≥
�π2
�2.
Mihaly Bencze
PP30426. Prove that�cos π
7 − cos 2π7 + cos 3π
7
�3= cos π
7 cos2π7 cos 3π
7 .
Mihaly Bencze
PP30427. In all tetrahedron ABCD holds32
√2R3
27 ≥ (P
rA)�P
r2A�≥ 32
√2r3.
Mihaly Bencze
PP30428. Determine all a, b ∈ R for which∞R0
�1
a+x2 − b cosx�
dxx = C when
C is the Euler constant.
Mihaly Bencze
PP30429. Determine all ak > 0 (k = 1, 2, ..., n) for which the function
f : (0,+∞) → R when f (x) =
�1n
nPk=1
axk
� 1x
is log− concave.
Mihaly Bencze
PP30430. Determine all z, v, w ∈ C for which |z| = |v| = |w| = 1 and��1 + z + v2 + w3�� ≥ 1;
��1 + v + w2 + z3�� ≥ 1;
��1 + w + z2 + v3�� ≥ 1.
Mihaly Bencze
PP30431. Let ABCD be a tetrahedron, denote r1, r2, r3, r4 the radii ofinspheres of tetrahedron GABC, GBCD, GCDA, GDAB where G is thecentroid of the given tetrahedron. Prove that16r3
R2
�1r1
+ 1r2
+ 1r3
+ 1r4
�≤ R2
r3(r1 + r2 + r3 + r4) .
Mihaly Bencze
1132 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP30432. Denote P1, P2, ..., Pk, ... a sequence of convex polygons such thatfor all k ≥ 1 the edges of plygon Pk+1 dividet the sides of polygon Pk in sameraport. Prove that exist only one point in interior of all polygons.
Mihaly Bencze
PP30433. Determine all n, k ∈ N ∗ for which
tg nπ2k+1 tg
kπ2n+1 ≥
�π2 + 2tg (n−1)π
2k+1
��π2 + 2tg (k−1)π
2n+1
�.
Mihaly Bencze and Gyorgy Szollosy
PP30434. In all triangle ABC holdsQ(2λa+ b+ c) ≥ 2 (2λ+ 1)2 s
�s2 + r2 + 2Rr
�,
Q(2λra + rb + rc) ≥ 4 (2λ+ 1)2 s2R for all λ ≥ 1+
√5
2 .
Mihaly Bencze
PP30435. Let ABC be a triangle and m ≤ A,B,C ≤ M. Prove thatPtg
�πA2
2(A2+1)
�≤ (M+m)2
4mM .
Mihaly Bencze
PP30436. Prove that∞Pk=1
tg�
π2(k2+1)
�< π2
6 .
Mihaly Bencze
PP30437. If 0 < a < b < 1 thenbRa
arctgxdxx > π
4 lnb2+1a2+1
.
Mihaly Bencze
PP30438. If b > a > 1 thenbRa
arctgxdxx < π
4 lnb2+1a2+1
.
Mihaly Bencze
PP30439. If b > a > 0 thenbRaarctgx · arctg 1
xdx < π4 ln
b2+1a2+1
.
Mihaly Bencze
Proposed Problems 1133
PP30440. Prove thatnP
k=1
1√k2+k−1
arctg√k2 + k − 1arctg 1√
k2+k−1< nπ
2(n+1)
Mihaly Bencze
PP30441. In all triangle ABC holdsP �
A2 + 1�arctgAarctg 1
A < π2 .
Mihaly Bencze
PP30442. In all triangle ABC holdsP 1
arctgAarctg 1A
> 12π .
Mihaly Bencze
PP30443. If 0 < a < b then
(n+1)π2
bRa
�
x2
2+lnx
�n
dx
arctgxarctg 1x
>�b2
2 + ln b�n+1
−�a2
2 + ln a�n+1
for all n ∈ N∗.
Mihaly Bencze
PP30444. In all triangle ABC holdsP
tg πA2A+π < π <
Ptg π2A
2(πA+2) .
Mihaly Bencze
PP30445. In all acute triangle ABC holds4π
P Aπ−2A < 2sr
s2−(2R+r)2< π
P Aπ−2A .
Mihaly Bencze
PP30446. In all triangle ABC holds:P �A+ π
2
�arctgA > π2
2 >P �
A+ 2π
�arctgA.
Mihaly Bencze
PP30447. If 0 < a < b then π2 ln
2b+π2a+π <
bRa
arctgxx dx < π
2 lnπb+2πa+2 .
Mihaly Bencze
PP30448. If 0 < a < b then π4 ln
2b2+π2a2+π
<bRa
arctg(x2)dxx < π
2 lnπb2+2πa2+2
.
Mihaly Bencze
1134 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP30449. In all acute triangle ABC holds1).
P Aπ−A ≤ 4R+r
s ≤P πA2(π−A)
2).P AB
(π−A)(π−B) ≤ 1 ≤P π2AB4(π−A)(π−B)
3). ABC(π−A)(π−B)(π−C) ≤
rs ≤ π3ABC
8(π−A)(π−B)(π−C)
Mihaly Bencze
PP30450. If 0 < a < b < 1 then π2 ln
2b+π2a+π ≤
bRa
arctgxx dx ≤ π
2 lnb+1a+1 .
Mihaly Bencze
PP30451. If 0 < a < b < 1 then π4 ln
2b2+π2a2+π
≤bRa
arctg(x2)dxx ≤ π ln b2+1
a2+1.
Mihaly Bencze
PP30452. In all triangle ABC holdsP
tg πA2A+π ≤ π ≤P tg πA
2(A+1) .
Mihaly Bencze
PP30453. If 0 < a < b < 1 then
13 ln
(2b+π)(π−a)(2a+π)(π−b) ≤
bRa
actgxtg x2
x2 dx ≤ π2
4 ln (b+1)(π−a)(a+1)(π−b) .
Mihaly Bencze
PP30454. In all triangle ABC holds
Ptg
λA2+(λ2−1)AA2+λA+λ2−1
≤ 2λ ≤P tgλ(λ2−1)A2+A
(λ2−1)A2+λA+1when λ = π
2 .
Mihaly Bencze
PP30455. In all acute triangle ABC holds
1). 2P 1
A − 6π ≤ s
r ≤ πP 1
A − 32). 4
π2
P �πA − 1
� �πB − 1
�≤ 4R+r
r ≤P�πA − 1
� �πB − 1
�
3). 8π3
�πA − 1
� �πB − 1
� �πC − 1
�≤ s
r ≤�πA − 1
� �πB − 1
� �πC − 1
�
Mihaly Bencze
PP30456. In all acute triangle ABC holdsP
�
λA−1+√
(λA−1)2+4A(λ2−1)(λ−A)��
λB−1+√
(λB−1)2+4B(λ2−1)(λ−B)�
4(λ2−1)2(λ−A)(λ−B)≤
≤ s2−r2−4Rrs2−(2R+r)2
≤
Proposed Problems 1135
P�
λA+1−λ2+√
(λA+1−λ2)2+4A(λ2−1)(λ−A)��
λB+1−λ2+√
(λB+1−λ2)2+4B(λ2−1)(λ−B)�
4(λ−A)(λ−B) ,
where λ = π2 .
Mihaly Bencze
PP30457. In all acute triangle ABC holds
1).P λA−1+
√(λA−1)2+4A(λ2−1)(λ−A)2(λ2−1)(λ−A)
≤ 2srs2−(2R+r)2
≤
≤P λA+1−λ2+√
(λA+1−λ2)2+4A(λ2−1)(λ−A)2(λ−A) where λ = π
2 .
Mihaly Bencze
PP30458. Prove that the sequence xn = (−1)[3√n2+2017n+2018] where [·]
denote the integer part is not periodic.
Mihaly Bencze
PP30459. If a0 = a, a1 = b and a2 + b2 + c = kabc; an+2an = a2n+1 + c then
compute∞Pn=1
1a2n.
Mihaly Bencze
PP30460. In all acute triangle ABC holdsQ λA−1+
√(λA−1)2+4(λ2−1)(λ−A)2(λ2−1)(λ−A)
≤
≤ 2srs2−(2R+r)2
≤Q λA+1−λ2+√
(λA+1−λ2)2+4A(λ2−1)(λ−A)2(λ−A) where λ = π
2 .
Mihaly Bencze
PP30461. If a, b, c > 0 thenP a4+1
a3+1+P a4
b3+c3≥ 3
2 +P
a.
Mihaly Bencze
PP30462. Prove thatnP
k=1
1
kn�
11n
+ 12n
+...+ 1(k−1)n
+(n!)r−1
knr + 1(k+1)n
+...+ 1nn
� ≥ 1 for
all −n+1n−1 ≤ r < 1 and n ≥ 2, n ∈ N.
Mihaly Bencze
1136 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP30463. Determine all n, k ∈ N for which {n+ 1, n+ 2, ..., kn} can bepartitioned in four subsets A,B,C,D such the sum of element from A,B,C,Dare equals.
Mihaly Bencze
PP30464. Compute2πR0
sinx cos 2xdx(2−cos2 x)(2−cos2 2x)
dx.
Mihaly Bencze
PP30465. If λ = limn→∞
1n
nPk=1
2k−1√n2+8k2−8k+4
then compute
limn→∞
n
�λ− 1
n
nPk=1
2k−1√n2+8k2−8k+4
�.
Mihaly Bencze and Gyorgy Szollosy
PP30466. If a, b, c > 0 and abc = 1 thenP a3
a9+1≤ 3
2 .
Mihaly Bencze
PP30467. In all triangle ABC holdsP
(sinA)2k (cosA)2(n−k) ≤ 32(nk)
where
n ≥ 2, k ∈ {1, 2, ..., n− 1} .
Mihaly Bencze
PP30468. ComputenP
k=1
n�
1√π− (2n−1)!!
(2n)!!
qn+ 1
2
�.
Mihaly Bencze
PP30469. If a, b, c ≥ 1 thenP 1
a+1 ≥P 1√ab+1
+ 16
�Pr √ab−1
(1+a)(1+b)(1+√ab)
���√a−
√b����2
.
Mihaly Bencze
PP30470. If ai > 0 (i = 1, 2, ..., n) andnQ
i=1ak = 1 then
nPi=1
11+ai+a2i+...+aki
≥ nk+1 , where k ∈ N∗.
Mihaly Bencze
Proposed Problems 1137
PP30471. In all triangle ABC holds
P�2
a3+1+ 1
b6+1
�≤ 3
2
��s2+r2+4Rr
4sRr
�2− 1
Rr
�.
Mihaly Bencze
PP30472. In all triangle ABC holdsP�
2r3a+1
+ 1r6b+1
�≤ 3(s2−8Rr−2r2)
2s2r2.
Mihaly Bencze
PP30473. Determine all λ > 0 such thatλR0
�1x + 1
ln(λ−x)
�dx = C, where C
is the Euler constant.
Mihaly Bencze
PP30474. Prove that 2nP
k=1
1
1+(k(k+1))32+
nPk=1
11+k3(k+1)3
≤ 3n2(n+1) .
Mihaly Bencze
PP30475. In all triangle ABC holds 2P
ln�A3 + 1
�+P
arctg�A3
�≤ 9π
2 .
Mihaly Bencze
PP30476. In all triangle ABC holds2P
ln�1 + r3a
�+P
arctg�r3a�≤ 9(4R+r)
2 .
Mihaly Bencze
PP30477. In all triangle ABC holdsP�
2
1+(sin A2 )
6 + 1
1+(sin B2 )
12
�≤ 3(s2+r2−8Rr)
2r2.
Mihaly Bencze
PP30478. In all triangle ABC holdsP�
2
1+(cos A2 )
6 + 1
1+(cos B2 )
12
�≤ 3
2
�1 +
�4R+r
s
�2�.
Mihaly Bencze
1138 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP30479. In all triangle ABC holdsP 1
2 ln�
1+(cos A2 )
6�
+arctg�
(cos A2 )
6� ≥ 2
9
�1 +
�4R+r
s
�2�.
Mihaly Bencze
PP30480. If 0 < a < b then 2 ln b3+1a3+1
+ arctg b3−a3
1+b3a3≤ 9
2 (b− a) .
Mihaly Bencze
PP30481. In all triangle ABC holdsP 1
2 ln(1+h3a)+arctg(h3
a)≥ 2r
9 .
Mihaly Bencze
PP30482. In all triangle ABC holdsP 1
2 ln�
1+(sin A2 )
6�
+arctg�
(sin A2 )
6� ≥ 2(s2+r2−8Rr)
9r2.
Mihaly Bencze
PP30483. Solve on (0,+∞) the following system:2x2
1
x32+1
+x23
x64+1
=2x2
2
x33+1
+x24
x65+1
= ... = 2x2n
x31+1
+x22
x63+1
= 32 .
Mihaly Bencze
PP30484. If x, y ≥ 0 then�2 ln
�x3 + 1
�+ arctg
�y3�� �
2 ln�y3 + 1
�+ arctg
�x3
��≤ 81
4 xy.
Mihaly Bencze
PP30485. Solve in (0,+∞) the following system:
3�x91 + x62 + x33 + 1
�= 2
�2x84 + x55 + 3x26
�
3�x92 + x63 + x34 + 1
�= 2
�2x85 + x56 + 3x27
�
−−−−−−−−−−−−−−−−−−3�x9n + x61 + x32 + 1
�= 2
�2x83 + x54 + 3x25
�.
Mihaly Bencze
PP30486. If a1 = 1 and (n+1)!an+1
= n!an
+ 2nn4+n2+1
for all n ≥ 1 then compute
limn→∞
1n3·n!
nPk=1
ak.
Mihaly Bencze and Lajos Longaver
Proposed Problems 1139
PP30487. If x1 ≥ 1 and xn+1 = 1 + x1x2...xn for all n ≥ 1 then1x3n+ 1
(1−xn)3 + 1
(xn+1−1)3= 3
xn(1−xn)(xn+1−1) for all n ≥ 1.
Mihaly Bencze
PP30488. If x1 ≥ 1 and xn+1 = 1 + x1x2...xn for all n ≥ 1 then compute
limn→∞
n
�2x1
−nP
k=1
1xk
�.
Mihaly Bencze
PP30489. If x ∈�0, π2
�then�
(sinx)8 + (cosx)8 + 1��
1(sinx)4
+ 1(cosx)4
+ 1�2
≥ 7298 .
Mihaly Bencze
PP30490. Denote x1 and x2 the roots of the equation(2m+ 1)x2 −
�m2 − 1
�x−m2 − 2m− 2 = 0. Determine all m, α ∈ R for
which 1|x1| ≤ sinα and 1
|x2| ≤ cosα.
Mihaly Bencze and Lajos Longaver
PP30491. Prove thatnP
k=1
(k+1)3
k(k+2)4≥ n
3(n+3) .
Mihaly Bencze
PP30492. If 0 < a ≤ b then
1). 3 ln b+3a+3 +
bRa
�x+1x+2
�3dx ≥ b− a
2). b− a+ 3 ln ba ≥
bRa
�x+2x+1
�3dx.
Mihaly Bencze
PP30493. Prove that 32arctg
2(b−a)4+ab +
bRa
�x2+2x2+3
�3dx ≥ b− a.
Mihaly Bencze
PP30494. Determine all x, y > 0 for which�x+1y+2
�3+�y+1x+2
�3≥ x
y+3 + yx+3 .
Mihaly Bencze
1140 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP30495. Compute limn→∞
n
�1√e−�
nPk=1
1√n2+k
�n�.
Mihaly Bencze
PP30496. Solve the following system:
x21 + x1x2 + x22 = 3x3 − x4−−−−−−−−−−−x22 + x2x3 + x23 = 3x4 − x5x2n + xnx1 + x21 = 3x2 − x3
.
Mihaly Bencze
PP30497. Compute∞Pk=1
1
(4k−1)k2.
Mihaly Bencze
PP30498. Determine all a, b, c, d ∈ N for which all numbers2a + 3b + 5c + 7d; 2b + 3c + 5d + 7a; 2c + 3d + 5a + 7b; 2d + 3a + 5b + 7c aredivisible by 17.
Mihaly Bencze
PP30499. Determine all matrices Ak ∈ M2 (C) (k = 1, 2, ..., n) for which
P1≤i<j≤n
(AiAj)−1 =
P
1≤i<j≤nAiAj
!−1
.
Mihaly Bencze
PP30500. Prove thatmPk=1
sin πk(k+1)n ≥ m
m+1 sinπn where m,n ∈ N∗.
Mihaly Bencze
PP30501. Determine all xk ∈ C (k = 1, 2, ..., n) for whichnP
k=1
xk = 1,
nQk=1
xk = 1 and |xk| = 1 (k = 1, 2, ..., n) .
Mihaly Bencze
PP30502. In all acute triangle ABC holds sr�s2 − (2R+ r)2
�≤ R4.
Mihaly Bencze
Proposed Problems 1141
PP30503. If a > 1 and G = (0,+∞),
x ∗ y = a loga
�1 +
�a
xa − 1
��a
ya − 1
��for all x, y ∈ G then (G, ∗) is abelian
group.
Mihaly Bencze
PP30504. Solve in R the following system:
x = y + 2 {z}y = z + 3 {x}z = x+ 4 {y}
where {·}
denote the fractional part.
Mihaly Bencze
PP30505. Solve in R the following system:
x2 = yz − 2 (2x+ z) {x}y2 = xz + {y}+ {z}z2 = xy + 2 (2y + z) {x}
where {·} denote the fractional part.
Mihaly Bencze
PP30506. Let ABC be a triangle and A1 ∈ (BC) , B1 ∈ (CA) , C1 ∈ (AB) .Determine the geometrical locus of points A1, B1, C1 for whichArea (A1B1C1) =
14Area (ABC) .
Mihaly Bencze
PP30507. Solve in R the following system
loga x+ loga+1 (y + 1) = loga+2
�z2 + 4x+ 4
�
loga y + loga+1 (z + 1) = loga+2
�x2 + 4y + 4
�
loga z + loga+1 (x+ 1) = loga+2
�y2 + 4z + 4
� where a > 1.
Mihaly Bencze
PP30508. If n ≥ 3, zk ∈ C (k = 1, 2, ..., n), zi 6= zj (i 6= j) then determine
all r ∈ N for which exist z ∈ C such that |z| = 1 andnP
k=1
|zr − zrk| ≥ n.
Mihaly Bencze
PP30509. Prove that exist a nonconstant geometrical progression formedby n natural numbers and which product is perfect n+ 1 power of a naturalnumber.
Mihaly Bencze
1142 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP30510. Prove that exist a nonconstant arithmetical progression formedby n natural numbers and which product is a perfect n+ 1 power of anatural number.
Mihaly Bencze
PP30511. Solve in R the following system:
52x+1 + (7y − 29) 5z + 3z2 − 17x = 152y+1 + (7z − 29) 5x + 3x2 − 17y = 152z+1 + (7x− 29) 5y + 3y2 − 17z = 1
.
Mihaly Bencze
PP30512. Prove that for all k ∈ N, k ≥ 2 exist points A,B,C on the graphicof the function f : (0,+∞) → R, f (x) = x2n+1 for which Area [ABC] = k2n.
Mihaly Bencze
PP30513. Compute limn→∞
n
e2
2 − n1+ 1
nR
−(1+ 1n)
x2n
1+ex dx
.
Mihaly Bencze
PP30514. If ak, bk, ck ∈ (0, 1) ∪ (1,+∞) (k = 1, 2, ..., n) then
3
snP
k=1
�logak bk
�3+ 3
snP
k=1
�logbk ck
�3+ 3
snP
k=1
�logck ak
�3 ≥ 3 3√n.
Mihaly Bencze
PP30515. Solve on R the equation(9x + 16x + 25x) (3−x + 4−x + 5−x) = 5
564 (3x + 4x + 5x)2 .
Mihaly Bencze
PP30516. Determine all ak ∈ (0,+∞) (k = 1, 2, ..., n) for which holds
[x] +n−1Pk=1
[x+ ak] = [anx] for all x ∈ R, where [·] denote the integer part.
Mihaly Bencze and Nicolae Papacu
Proposed Problems 1143
PP30517. Compute limn→∞
nR0
dx
(1+x+x2+...+xk)(1+xn)where k ∈ {1, 2, ..., n} .
Mihaly Bencze
PP30518. If x, y, z > 0 andP x
y = 27 thenP x
√x√
x3+xy2+2y3≥
√32 .
Mihaly Bencze
PP30519. Solve in Z the equation x2
2 + y3
3 + z4
4 = (z − 1) (x+ y)2 .
Mihaly Bencze
PP30520. If mk ∈ N∗ (k = 1, 2, ..., n) then
�n
nQk=1
mk
�sin π
n�
k=1mk
≥nP
k=1
mk sinπmk
.
Mihaly Bencze
PP30521. Prove thatnP
k=1
sin πmk
sin πk
≥ 2nm(n+1) where n,m ∈ N∗.
Mihaly Bencze
PP30522. Compute limn→∞
1n
nPk=1
π2R0
cosx√k+sin 2x
dx.
Mihaly Bencze
PP30523. Let ABC be a triangle in which1
BC+λAB + 1BC+λAC = 1
λAB−BC + 1λAC−BC .
1). Determine all λ ∈ R for which ABC is rectangle2). Determine all λ ∈ R for which ABC is isoscele
Mihaly Bencze and Sandor Tamas
PP30524. If xk > 0 (k = 1, 2, ..., n) andnQ
k=1
xk ≤ 2n thennP
k=1
11+xk
≥ 1.
Mihaly Bencze
1144 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018
PP30525. If the equation x2 − ax+ b = 0 have integer roots, where a, b ∈ Rthen b3 + 2a3 − 6ab+ 9 and b4 + 4a4 − 16a2b+ 8b2 + 16 are compositenumbers.
Mihaly Bencze
PP30526. Let be Mn =nx ∈ R| [x]{x} = n ∈ N∗
owhere [·] and {·} denote the
integer respective fraction part. Compute∞Pk=1
1x21+x2
2+...+x2nk
where
x1, x2, ..., xnk∈ Mn.
Mihaly Bencze
PP30527. Determine all X ∈ M3 (R) for which xn + xn+1 =
2 2 20 2 20 0 2
where n ∈ N.
Mihaly Bencze
PP30528. If x1 ≥ 1 and xn+1 = 1 +nQ
k=1
xk for all n ≥ 1 then compute
∞Pn=1
1x2n.
Mihaly Bencze
PP30529. In all triangle ABC holdsP (ra+rb)
5+5r5c(ra+rb+4rc)r4c
≥ 32 + 32Rr .
Mihaly Bencze
PP30530. In all triangle ABC holdsP r10a +5r5
br5c
(r2a+2rbrc)r5br5c
≥ 3 + (4R+r)3
s2r− 12R
r .
Mihaly Bencze
PP30531. In all triangle ABC holdsP (s−a)10+5(s−b)5(s−c)5
((s−a)2+2(s−b)(s−c))(s−b)4(s−c)4≥ 3 + s2−12Rr
r2.
Mihaly Bencze
Proposed Problems 1145
PP30532. In all triangle ABC holdsP (ha+hb)
5+160h5c
(ha+hb+4hc)h4c≥ 40 +
4(s2+r2)Rr .
Mihaly Bencze
PP30533. In all triangle ABC holdsP (a+b)5+160c5
(a+b+4c)c4≥ 40 +
4(s2+r2)Rr .
Mihaly Bencze
PP30534. In all triangle ABC holdsP a10+5b5c5
(a2+2bc)b4c4≥ s2−3r2
2Rr .
Mihaly Bencze
PP30535. In all triangle ABC holdsP (sin2 A
2+sin2 B
2 )5+160 sin10 C
2
(sin2 A2+sin2 B
2+2 sin2 C
2 ) sin8 C
2
≥ 96 +8((2R−r)2(s2+r2−8Rr)−2Rr2)
Rr2.
Mihaly Bencze
PP30536. In all triangle ABC holdsP (cos2 A
2+cos2 B
2 )5+160 cos10 C
2
(cos2 A2+cos2 B
2+2 cos2 C
2 ) cos8C2
≥ 96 +4((4R+r)3+s2(2R+r))
Rs2.
Mihaly Bencze
PP30537. In all acute triangle ABC holdsP (cosA+cosB)5+160 cos5 C
(cosA+cosB+2 cosC) cos4 C≥ 96 +
16((2R+r)2+s2r−2R(s2+r2+2Rr))R(s2−(2R+r)2)
.
Mihaly Bencze
PP30538. In all acute triangle ABC holdsP cos10 A+5 cos5 B cos5 C
(cos2 A+cosB cosC) cos4 B cos4 C≥ 9 +
3(s2+r2−4R2)((4R+r)2−3s2)16R4 .
Mihaly Bencze
PP30539. In all triangle ABC holdsP (ha+hb)
5+160h5c
(ha+hb+2hc)h4c≥ 80 +
8(s2+r2)Rr .
Mihaly Bencze
PP30540. In all triangle ABC holdsP (ra+rb)
5+160r5c(ra+rb+2rc)
≥ 80 + 32Rr .
Mihaly Bencze